AUG  1 


2  9 


1950 


Southern  Branch 
of  the 

University  of  California 


Los  Angeles 


Form  L   1 


This  book  is  DUE  on  the  last  date  stamped  below 

JUL  8      1*2? 


JUL  1  8  W 


I  2 


2 


7     1 
8 


'JUL  2  9  1953 
MAN  2  7  1954 


1962 
-  2  3  1979 


n  L-9-15m-8,'26 


GENERAL  MATHEMATICS 


BY 

RALEIGH  SCHORLING 

IN  CHARGE  OF  MATHEMATICS,  THE   LINCOLN  SCHOOL  OF  TEACHERS 
COLLEGE,  NEW  YORK  CITY 


AND 


WILLIAM  DAVID  REEVE 

TEACHERS'   TRAINING   COURSE  IN   MATHEMATICS  IN   THE   COLLEGE 
OF  EDUCATION,  ANI>  HEAD  OF  THE  MATHEMATICS  DEPART- 
MENT   IN  THE    UNIVERSITY   HIGH   SCHOOL 
THE   UNIVERSITY  OF  MINNESOTA 


GINN  AND  COMPANY 

BOSTON    •    NEW  YORK    •    CHICAGO    •     LONDON 
ATLANTA    •    DALLAS     •    COLUMBUS    •    SAN   FRANCISCO 


2883 


171-50 

m 


COPYRIGHT,  1919,  BY 
RALEIGH  SCHORLING  AND  WILLIAM  DAVID  REEVE 


ALL,  RIGHTS   RESERVED 
819.11 


7171 


gtbenaum 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


•''A- 


PREFACE 

The  purpose  of  this  book,  as  implied  in  the  introduction, 
is  as  follows :  to  obtain  a  vital,  modern  scholarly  course  in 
introductory  mathematics  that  may  serve  to  give  such  careful 
training  in  quantitative  thinking  and  expression  as  well- 
informed  citizens  of  a  democracy  should  possess.  It  is,  of 
course,  not  asserted  that  this  ideal  has  been  attained.  Our 
achievements  are  not  the  measure  of  our  desires  to  im- 
prove the  situation.  There  is  still  a  very  large  "  safety 
factor  of  deud  wood"  in  this  text.  The  material  purposes 
to  present  such  simple  and  significant  principles  of  algebra, 
geometry,  trigonometry,  practical  drawing,  and  statistics, 
along  with  a  few  elementary  notions  of  other  mathematical 
subjects,  the  whole  involving  numerous  and  rigorous  appli- 
cations of  arithmetic,  as  the  average  man  (more  accurately 
the  modal  man)  is  likely  to  remember  and  to  use.  There 
is  here  an  attempt  to  teach  pupils  things  worth  knowing 
and  to  discipline  them  rigorously  in  things  worth  doing. 

The  argument  for  a  thorough  reorganization  need  not 
be  stated  here  in  great  detail.  But  it  will  be  helpful  to 
enumerate  some  of  the  major  errors  of  secondary-mathe- 
matics instruction  in  current  practice  and  to  indicate 
briefly  how  this  work  attempts  to  improve  the  situation. 
The  following  serve  to  illustrate  its  purpose  and  program : 
1.  The  conventional  first-year  algebra  course  is  charac- 
terized by  excessive  formalism;  and  there  is  much  drill 
work  largely  on  nonessentials.  The  excessive  formalism  is 


iv  GENERAL  MATHEMATICS 

greatly  reduced  in  this  text  and  the  emphasis  placed  on 
those  topics  concerning  which  there  is  general  agreement, 
namely,  function,  equation,  graph,  and  formula.  The  time 
thus  gained  permits  more  ample  illustrations  and  applica- 
tions of  principles  and  the  introduction  of  more  significant 
material. 

2.  Instead    of    crowding    the   many   difficulties    of   the 
traditional  geometry  course   into    one  year,  geometry  in- 
struction is  spread  over  the  years  that  precede  the  formal 
course,  and  the  relations  are  taught  inductively  by  experi- 
ment and  by  measurement.    Many  foreign  schools  and  an 
increasing  number  of  American  schools  proceed   on  this 
common-sense  basis.    This  gives  the  pupil  the  vocabulary, 
the  symbolism,  and  the  fundamental  ideas  of  geometry. 
If  the  pupil  leaves  school  or  drops  mathematics,  he  never- 
theless has  an  effective  organization  of  geometric  relations. 
On  the  other  hand,  if  he  later  pursues  a  formal  geometry 
course,  he  can  work  far  more  effectively  because  he  can 
concentrate  on  the  logical  organization  of  space  relations 
and  the  formal  expression  of  these  relations.    The  longer 
"  time  exposure "  minimizes  the  difficulties  met  in  begin- 
ning  the   traditional   geometry    courses    and    avoids    the 
serious  mistake  of  forcing  deductive  logic  and  philosophic 
criticism  in  these  early  years. 

3.  The  traditional    courses  delay  the  consideration    of 
much  interesting  and  valuable  material  that  the  field  of 
secondary  mathematics  has  to  offer,  and  which  may  well 
be  used  to   give  the  pupil  very  early  an   idea  of  what 
mathematics  means  and  something  of  the  wonderful  scope 
of  its  application.   The  material  of  the  seventh,  eighth,  and 
ninth  years  is  often  indefensibly  meaningless  when  com- 
pared with  that  of  many  foreign  curricula.    Trigonometry, 


containing  many  easy  real  problems,  furnishes  a  good  ex- 
ample of  this  delay.  Other  examples  are  found  in  the 
use  of  logarithms,  the  slide  rule,  standardized  graphical 
methods,  the  notion  of  function,  the  common  construction 
of  practical  drawing,  the  motivation  of  precise  measure- 
ment, a  study  of  the  importance  of  measurement  in  modern 
life,  and  the  introductory  ideas  of  the  calculus.  It  appears 
that  the  mathematics  student  should  be  given  an  oppor- 
tunity to  use  these  important  tools  very  early  in  his  study. 
They  lend  to  the  subject  a  power  and  interest  that  drills  on 
formal  material  cannot  possibly  give. 

Particular  emphasis  is  given  to  graphical  representation 
of  statistics.  The  growing  complexity  of  our  social  life 
makes  it  necessary  that  the  intelligent  general  reader 
possess  elementary  notions  of  statistical  methods.  The 
hundreds  of  articles  in  the  current  magazines  so  exten- 
sively read  demand  an  elementary  knowledge  of  these 
things  in  order  that  the  pupil  may  not  remain  ignorant  of 
the  common,  everyday  things  of  life.  Brief  chapters  on 
logarithms  and  the  slide  rule  have  been  introduced  in 
order  that  a  greater  number  of  students  may  use  these 
practical  labor-saving  devices  and  in  order  that  these  devices 
may  function  in  the  student's  subsequent  work,  whether 
in  everyday  life  or  in  the  classroom.  Actual  classroom 
experience  with  these  chapters  has  proved  them  to  be 
relatively  simple  and  good  material  for  eighth-grade  and 
ninth-grade  students. 

4.  Mathematics  needs  to  be  reorganized  on  the  side 
of  method.  The  information  we  now  possess  of  individual 
differences  and  effective  devices  in  supervised  study  should 
make  the  study  of  mathematics  more  nearly  a  laboratory 
course,  in  which  more  effective  work  can  be  done. 


vi  ( i  KN  KJ{  A  L   M  ATI  1  KM  AT1CS 

5.  The  teaching  of  algebra,  geometry,  and  trigonometry 
in  separate  fields  is  an  artificial  arrangement  that  does  not 
permit  the  easy  solution  of  problems  concerning  projects 
that  correlate  with  problems  met  in  the  physical  and  bio- 
logical sciences  or  the  manual  and  fine  arts.    To  reject  the 
formalism  of  algebra,  to  delay  the  demands  of  a  logical 
unit  in  geometry,  and  to  present  the  simple  principles  of 
the  various  branches  of  mathematics  in  the  introductory 
course  opens  the  door  to  a  greater  variety  of  problems  that 
seem  to  be  real  applications.    The  pupil  sees  the  usefulness 
of  the  various  modes  of  treatment  of  the  facts  of  quantity. 
Power  is  gained  because  the  pupil  is  equipped  with  more 
tools,  in  that  the  method  of  attack  is  not  limited  to  one  field. 

6.  One  of  the  most  curious  characteristics  of  American 
secondary-mathematics  instruction  is  the  obscurity  in  the 
teaching   of  the   function  notion.    It  is  generally  agreed 
that  functional  thinking  (the  dependence  of  one  magnitude 
upon  another)  constitutes  one  of  the   most  fundamental 
notions  of  mathematics.    Because  of  the  interrelations  of 
the  equation,  the  formula,  the  function,  the  graph,  and  the 
geometric  relations  inductively   acquired,   the   material  is 
easily  correlated  around  the  function  idea  as  the  organizing 
and  unifying  principle.   The  function  concept  (implicitly  or 
explicitly)  dominant  throughout  helps  to  lend  concreteness 
and  coherence  to  the  subject.    However,  it  would  be  false 
to  assume  that  this  material  is  presented  to  establish  the 
principle  of  correlation.    On  the  contrary,  it  happens  that 
correlation  around  the  function  notion,  though  incidental, 
is  a  valuable  instrument  for  accomplishing  the  larger  aim, 
which  is  to  obtain  a  composite  introductory  course  in  mathe- 
matics that  all  future  citizens  of  our  democracy  should  be 
required  to  take  as  a  matter  of  general  scholarship. 


PREFACE  vii 

7.  The  traditional  reticence  of  texts  has  made  mathe- 
matics unnecessarily  difficult  for  pupils  in  the  early  years. 
The  style  of  this  book,  though  less  rigidly  mathematical,  is 
more  nearly  adapted  to  the  pupils'  mental  age.  The  result 
is  a  misleading  length  of  the  book.  The  book  can  easily  be 
taught  in  a  school  year  of  approximately  one  hundred  and 
sixty  recitations.  In  the  typical  high  school  it  will  be  taught 
in  the  first  year.  (The  Minnesota  high  schools  taught  it 
in  this  grade,  five  recitations  per  week.)  In  schools  which 
control  the  seventh  and  eighth  years  the  following  are 
also  possibilities  which  have  been  tested  by  the  authors 
and  cooperating  teachers :  (1)  in  the  eighth  year  with  daily 
recitations ;  (2)  half  of  the  book  in  the  eighth  year  and  the 
remainder  in  the  ninth  year,  with  three  recitations  per  week 
(it  was  so  used  in  the  Lincoln  School) ;  (3)  the  course 
may  be  started  in  the  seventh  year  provided  the  class  has 
achieved  good  results  in  previous  arithmetic  work. 

Specific  references  are  given  where  material  which  is 
not  of  the  common  stock  has  been  taken  consciously,  the 
purpose,  however,  being  chiefly  to  stimulate  pupils  and 
teachers  to  become  familiar  with  these  books  for  reasons 
other  than  the  obligations  involved.  Something  of  human 
interest  is  added  by  relating  some  of  the  well-known  stories 
of  great  mathematicians.  We  are  indebted  to  Professor 
David  Eugene  Smith  on  questions  relating  to  historical 
material.  In  our  thinking  we  are  particularly  indebted  to 
Professors  Nunn,  Smith,  Breslich,  and  Myers.  We  shall 
be  obliged  to  all  teachers  who  may  think  it  worth  while 

to  point  out  such  errors  as  still  exist. 

THE  AUTHORS 


CONTENTS 

CHAPTER  PAGE 

I.  THE  EQUATION 1 

Solving  an  equation 6 

Translation  of  an  equation .     .  12 

Solution  of  verbal  problems 16 

Axioms        21 

II.  LINEAR    MEASUREMENT.     THE    EQUATION    APPLIED    TO 

LENGTH 26 

Different  units  of  length 28 

Squared  paper 32 

Sum  of  two  segments ;  geometric  addition 36 

Polygons * 44 

III.  PROPERTIES  OF  ANGLES 47 

Notation  for  reading  angles 50 

Measurement  of  angles ;  the  protractor 54 

Measuring  angles  ;  drawing  angles 56 

Comparison  of  angles -  59 

Geometric  addition  and  subtraction  of  angles 61 

Parallel  lines 68 

How  to  construct  a  parallelogram 70 

IV.  THE  EQUATION  APPLIED  TO  AREA 74 

Formula 78 

Formula  for  the  area  of  a  parallelogram 79 

Geometric  interpretation  of  products 85 

Algebraic  multiplication 89 

The  accuracy  of  the  result 93 

V.  THE  EQUATION  APPLIED  TO  VOLUME 98 

Measurement  of  volume 99 

Formula  for  the  volume  of  a  rectangular  parallelepiped     .  99 

ix 


x  GE^'EKAL  MATHEMATICS 

CHAPTER  PAGE 

Formula  for  the  volume  of  a  cube 102 

Exponents 102 

Application  of  algebraic  principles  to  geometric  figures  .  105 

VJ.  THE    EQUATION   APPLIED   TO  FUNDAMENTAL  ANGLE 

RELATIONS Ill 

The  sum  of  all  the  angles  about  a  point  on  one  side 

of  a  straight  line 112 

Sum  of  all  the  angles  about  a  point  in  a  plane       .     .     .  113 

Supplementary  angles ;  supplement 116 

Complementary  angles 119 

Vertical  angles 122 

Important  theorems  relating  to  parallel  lines    ....  126 

VII.  THE  EQUATION  APPLIED  TO  THE  TRIANGLE  ....  130 

The  sum  of  the  interior  angles 131 

Right  triangle -.     .  135 

Exterior  angles  of  a  triangle 139 

The  construction  of  triangles 142 

VIII.  POSITIVE  AND  NEGATIVE  NUMBERS.    ADDITION  AND 

SUBTRACTION , 150 

Use  of  signs 151 

Geometric  representation  of  positive  numbers.    Origin  .  152 

Geometric  representation  of  negative  numbers      .     .     .  153 

Algebraic  addition 162 

Subtraction  illustrated  by  the  number  scale      ....  170 

Algebraic  subtraction 171 

Subtraction  of  polynomials 173 

IX.  POSITIVE  AND  NEGATIVE  NUMBERS.  MULTIPLICATION 

AND  DIVISION.    FACTORING 178 

Geometric  illustration  of  law  of  signs 178 

Law  of  signs  illustrated  by  a  balanced  bar 180 

Multiplication  of  positive  and  negative  numbers  .     .     .  182 

Special  products 192 

Law  of  signs  in  division 195 

Factoring 198 


CONTENTS  xi 

CHAPTER  PAGE 

Distinction  between  identity  and  equation 204 

Use  of  factoring  in  identities  for  calculating  areas     .     .  205 

X.  GRAPHICAL    REPRESENTATION    OF    STATISTICS;    THE 

GRAPH  OF  A  LINEAR  EQUATION 214 

Pictograms 214 

Practice  in  interpreting  the  bar  diagram 222 

How  to  construct  a  bar  diagram 224  ' 

Practice  in  interpreting  graphic  curves 231 

How  the. graphic  curve  is  drawn 233 

Normal  distribution 257 

Symmetry  of  a  curve 259 

Graph  of  constant  cost  relations 262 

Graphs  of  linear  equations 263 

XI.  GAINING    CONTROL    OF    THE    FORMULA;  GRAPHICAL 

INTERPRETATION  OF  FORMULAS 273 

Solving  a  formula 276 

Graphical  illustration  of  a  motion  problem 283 

Translating  rules  of  procedure  into  formulas    ....  287 

Graph  of  the  centigrade-Fahrenheit  formula     ....  288 

Evaluating  a  formula 290 

XII.  FUNCTION 299 

Graph  of  a  function 301 

Solving  the  function  set  equal  to  zero 304 

Direct  variation 305 

Graphing  direct  variation 308 

Graphing  inverse  variation    .     .     .     ...     .     .     .     .     .  311 

XIII.  SIMILARITY;  CONSTRUCTION  OF  SIMILAR  TRIANGLES  .  314 

Summary  of  constructions  for  similar  triangles     .     .     .  317 

Algebraic  problems  on  similar  figures      ......  319 

Proportion 322 

Construction  of  a  mean  proportional 332 

Fourth  proportional  construction 334 

Verbal  problems  solved  by  proportion 336 

Proportionality  of  areas 341 


xii  GENERAL  MATHEMATICS 

CHAPTER  PAGE 

XIV.  INDIRECT  MEASUREMENT;  SCALE  DRAWINGS;  TRIG- 
ONOMETRY       345 

Similar  right  triangles :;.",:, 

Trigonometric  ratios :;f>!t 

Table  of  trigonometric  ratios 361 

Verbal  trigonometry  problems 362 

XV.  THEORY     AND     APPLICATION     OF     SIMULTANEOUS 

LINEAR  EQUATIONS 367 

Graphic  solution 369 

Algebraic  methods  for  solving    simultaneous   linear 

equations 373 

Summary  of  methods  of  elimination 379 

Classified  verbal  problems 384 

XVI.  GEOMETRIC   AND   ALGEBRAIC    INTERPRETATION    OF 

ROOTS   AND   POWERS 390 

The  theorem  of  Pythagoras 397 

Constructing  the  square  root  of  a  number      ....  404 
Fractional   exponents    another    means   of   indicating 

roots  and  powers 412 

XVII.  LOGARITHMS *  ....  424 

• 

Logarithms  defined , 427 

Exponential  equations 443 

Interest  problems  solved  by  logarithms 444 

XVIII.  THE  SLIDE  RULE .  449 

Square  roots  found  by  means  of  the  slide  rule    .     .     .  455 

Verbal  problems  solved  by  the  slide  rule 458 

XIX.  QUADRATIC  FUNCTIONS;  QUADRATIC  EQUATIONS     .  462 

How  to  solve  a  quadratic  equation  graphically   .     .     .  465 

The  parabola 467 

More  powerful  methods  of  solving  quadratic  equations  471 

Maxima  and  minima  algebraically  determined  .     .     .  479 

INDEX                                    481 


INTRODUCTION 

The  movement  to  provide  an  introductory  course  in  general 
mathematics  is  a  part  of  an  extensive  movement  toward  mak- 
ing the  materials  of  study  in  secondary  education  more  concrete 
and  serviceable.  The  trend  of  education  expresses  a  determi- 
nation that  the  seventh,  eighth,  and  ninth  school  years  should 
be  enriched  by  the  introduction  of  such  significant  experiences 
of  science,  civics,  art,  and  other  knowledge  of  human  life  as 
all  enlightened  citizens  of  a  democracy  should  possess.  The 
work  of  these  grades  cannot  be  liberalized  by  "shoving  down" 
the  conventional  material  a  year  or  so.  The  reorganization 
must  be  more  fundamental  in  order  to  revitalize  and  socialize 
the  mathematics  of  these  grades. 

Competent  authorities  in  mathematics  have  from  time  to  time 
asserted,  first,  that  American  secondary-mathematics  teaching 
has  been  characterized  by  a  futile  attempt  to  induce  all  pupils 
to  become  technical  college  mathematicians.  Secondly,  that 
instead  of  giving  pupils  an  idea  of  the  real  meaning  of  mathe- 
matics and  the  wide  range  of  its  applications,  they  are  forced 
to  waste  a  great  deal  of  time  on  abstract  work  in  difficult 
problems  in  radicals,  fractions,  factoring,  quadratics,  and  the 
like,  which  do  not  lead  to  anything  important  in  mathematics. 
And,  thirdly,  that  this  meaningless  juggling  of  symbolism  fails 
to  meet  the  needs  of  the  great  number  of  pupils  who  go  rather 
early  into  their  careers ;  it  also  wastes  time  and  effort  on  the 
part  of  pupils  with  especial  ability  in  the  subject,  who  ought 
to  get  an  early  insight  into  the  scope  and  power  of  the  real 
science  of  mathematics. 


xiv  GENERAL  MATHEMATICS 

Quantitative  thinking  and  expression  play  so  large  a  part  in 
human  experience  that  proper  training  in  these  matters  will 
always  be  important.  The  growing  complexity  of  social  and 
industrial  life  is  responsible  for  corresponding  changes  in  the 
use  of  quantitative  relationships.  Old  applications'  in  many 
instances  are  disappearing,  but  new  ones  growing  out  of  present- 
day  relations  are  being  introduced  to  take  their  places.  These 
changes  require  a  new  kind  of  introductory  text  in  mathematics. 
Action  is  forced  by  the  demand  that  there  shall  be  justification 
of  the  time  and  effort  given  to  each  subject  and  each  item  in 
the  subject.  New  subjects  which  appear  necessary  in  the  proper- 
training  for  citizenship  are  crowding  the  curriculum.  Mathe- 
matics too  must  justify  its  place  "  in  the  sun  "  by  a  thorough 
reorganization  that  will  meet  modern  needs.  This  is  what  is 
meant  by  revitalizing  mathematics. 

The  practical  administrator  will  be  impressed  by  the  fact 
that  this  program  raises  no  administrative  difficulties.  The 
pupil  may  be  expected  to  develop  greater  power  in  algebra, 
because  the  elimination  of  material  which  wastes  time  and 
effort  has  made  possible  the  emphasizing  of  the  topics  concern- 
ing which  there  is  general  agreement.  The  supplementary 
material  which  is  drawn  from  the  other  subjects  constitutes  a 
preparation  for  further  study  in  these  fields ;  for  example,  the 
text  gives  the  pupil  the  vocabulary,  the  symbolism,  and  many 
of  the  ideas  of  plane  geometry. 

This  type  of  introductory  course  should  appeal  to  the  pro- 
gressive educator  because  of  a  number  of  other  features.  The 
"  problem  method"  of  teaching  is  followed  throughout.  Ration- 
alized drills  are  provided  in  abundance.  The  course  has  been 
used  in  mimeographed  form  by  experienced  teachers.  Scores 
of  prospective  teachers  have  found  the  treatment  simple  and 
easy  to  present.  Inexperienced  teachers  have  gone  out  into 
difficult  situations  and  have  taught  the  material  with  satis- 
faction. Pupils  following  this  course  have  made  better  progress 


INTRODUCTION  xv 

than  pupils  following  the  traditional  course,  and  both,  pupils 
and  teachers  manifest  a  degree  of  interest  seldom  seen  in  the 
ordinary  class  in  mathematics.  The  tests  prepared  by  the 
authors  will  save  time  for  teachers  and  enable  them,  if  they 
desire,  to  diagnose  their  own  situations  and  to  compare  their 
results  with  those  obtained  in  other  institutions  using  the  same 
material.  If  the  question  is  raised  as  to  what  students  com- 
pleting such  a  course  will  do  when  they  get  to  college,  it  may 
be  replied  that  enough  of  them  have  already  entered  college 
to  convince  the  unbiased  that  they  experience  no  handicap. 
The  more  important  point,  however,  is  that  such  a  course 
enables  one  to  understand  and  to  deal  with  the  quantitative 
world  in  which  he  lives. 

This  course  in  reorganized  introductory  mathematics, 
although  but  a  part  of  a  large  movement  in  secondary  educa- 
tion which  looks  toward  more  concrete  teaching  and  more 
serviceable  materials  of  study,  has  a  further  highly  significant 
aspect.  It  is  a  potent  and  encouraging  evidence  that  high- 
school  teachers  have  become  students  of  their  own  teaching, 
and  as  a  result  are  preparing  their  own  textbooks  in  the  midst 
of  real  teaching  situations,  as  the  outcome  of  intelligent  con- 
'structive  experimentation. 

Probably  very  few  books  have  been  subjected  previous  to 
publication  to  such  thorough  tests  of  teaching  situations.  The 
authors  have  been  shaping  this  course  for  many  years.  During 
the  last  three  years  the  manuscript  as  originally  accepted  by 
the  publishers  has  been  taught  in  mimeograph  form  to  more 
than  a  thousand  pupils  distributed  in  a  selection  of  typical 
schools,  among  these  being  the  following  :  Minneapolis  Central 
High  School  (large  city  high  school),  Bremer  Junior  High 
School,  Seward  Junior  High  School,  University  of  Minnesota 
High  School,  Owatonna  High  School,  Mabel  High  School 
(small  town),  and  the  Lincoln  School  of  Teachers'  College. 
Numerous  consultations  with  the  teachers  in  these  schools 


xvi  GENERAL  MATHEMATICS 

resulted  in  many  valuable  suggestions  which  contributed  directly 
toward  making  the  text  easily  teachable. 

Each  of  the  authors  has  taught  secondary  mathematics  for 
more  than  ten  years  in  large  public  and  private  schools.  They 
have,  supervised  many  teachers  in  training ;  they  have  taught 
teacher-training  courses,  and  each  during  most  of  this  time  has 
had  unusual  opportunities  for  free  experimentation.  The  text 
may  be  regarded  by  fellow  teachers  of  mathematics  as  a  report 
which  shows  in  organization  and  subject  matter  the  things  that 

have  seemed  most  useful. 

LOTUS  D.  COFFMAN 

OTIS  W.  CALDWELL 


FIRST  YEAR 


CHAPTER  I 

THE  EQUATION 

1.  A  problem  introducing  the  use  of  the  equation.    In 

order  to  find  the  weight  of  a  bag  of  candy,  it  was  placed 
on  one  pan  of  perfectly  balanced  scales  (Fig.  1).  The 
candy,  together  with  a  4-ounce  weight,  balanced  10  oz. 
of  weights  on  the 
other  pan.  How 
much  did  the  bag 
of  candy  weigh  ? 

It  is  a  familial- 
principle  of  bal- 
anced scales  that 
if  the  same  number 
of  ounces  be  taken 
from  each  pan,  the 
balance  is  not  disturbed.  Hence,  if  we  suppose  that  a 
4-ounce  weight  could  be  removed  from  each  pan,  the 
candy  would  be  balanced  by  6  oz. 

This  solves  the  problem,  but  let  us  analyze  it  a  little 
further.  The  important  fact  in  the  situation  above  is  that 
an  unknown  number  of  ounces  of  candy  plus  4  oz.  in 
one  pan  balances  10  oz.  in  the  other  pan.  If  we  agree 

1 


FIG.  1.  THE  BALANCED  SCALES  ILLUSTRATE 
THE  MEANING  OF  AN  EQUATION 


2  GENERAL  MATHEMATICS 

to  let  the  letter  w  represent  the  number  of  ounces  of 
weight  in  the  bag  of  candy  and  use  the  sign  of  equality 
(=)  to  denote  the  perfect  balance  of  the  scales,  the  pre- 
ceding mathematical  fact  may  be  conveniently  translated 
into  the  following  expression  :  w  +  4  =  10,  where  w  +  4 
denotes  the  weight  in  the  left  pan  and  10  the  weight  in 
the  right  pan.  The  abbreviated  ("shorthand")  statement, 
w  +  4=10,  expresses  equality  and  is  called  an  equation. 
The  number  to  the  left  side  of  the  equality  sign  is  called 
the  left  member  of  the  equation,  the  number  to  the  right 
is  the  right  member. 

Just  as  the  scales  will  balance  if  the  same  number  of 
ounces  are  taken  from  each  pan,  so  ive  may  subtract  the 
same  number  from  both  aides  of  an  equation  and  get  another 
f'/Hxtion.  In  the  preceding  problem  the  written  work  may 

be.  arranged  thus: 

f  number  of  ounces  of  weight 
Let  to  =  J.       .     ., 

^     in  the  bag  of  candy. 

Then  w  +  4  =  10 

4=4 
Subtracting  4  from  each  "I ^ 

member  of  the  equation,  J 
Thus,  the-  bag  of  candy  weighs  6  oz. 

The  preceding  problem  illustrates  the  principle  that  if 
the  same  number  be  subtracted  from  both  members  of  an 
equation,  the  remainders  are  equal ;  that  is,  another  equation 
is  obtained.  [Subtraction  Law] 

EXERCISES 

Find  the  value  of  the  unknown  numbers  in  the  following 
equations,  doing  all  you  can  orally : 

1.  x  +  2  =  6.  4.  or  + 11  =  18.  7.  x  +  10  =  27. 

2.  x  +  6  =  10.  5.  x  + 13  =  23.  8.  x  +  14  =  21. 

3.  x  +  7  =  l3.  6.  z+9  =  26.  9.  x  -f  33  =  44. 


THE  EQUATION 


3 


2.  The  importance  of  the  equation.    The  equation  is  a 
very  important  tool  for  solving  problems  in  the   mathe- 
matical sciences.    The  equation  gives  us  a  new  method  of 
attack  on  a  problem,  enabling  us  to  solve  many  problems 
which  would  be  very  difficult,  if  not  impossible,  without 
its  use. 

3.  Method  of  studying  the  nature  of  the  equation.    In 

making  a  study  of  the  equation  we  shall  continue  by  con- 
sidering some  very  simple  problems  in  order  that  we  may 
clearly  understand  the  laws  which  are  involved.  If  these 
laws  are  mastered  in  connection  with  the  simple  cases,  it 
will  be  easy  to  apply  the  equation  as  a  tool  for  solving 
the  more  complicated  and  difficult  problems.  In  the  next 
article  we  shall  continue  to  interpret  the  equation  by 
considering  a  problem  in  weighing. 

4.  Division  Law.   Two  equal  but  unknown  weights,  to- 
gether with  a  1-pound  weight,  just  balance  a  16-pound  and 
a  1-pound  weight  to- 
gether (Fig,  2).  How 

heavy    is    each    un- 
known weight? 

Let  p  equal  the 
number  of  pounds  in 
one  of  the  unknown 
weights.  Suppose  that 
1  Ib.  be  removed  from 
each  pan,  leaving  *2p  FIG.  2.  THE  BALANCED  SCALES  MAT  UK  USED 
DOUnds  in  the  left  T(?  ILLUSTRATE  THE  SUBTRACTION  LAW  AND 

THE  DIVISION  LAW  i 

pan  balancing  the  re- 
maining 16  Ib.  in  the    right   pan.     Then,  if  2  p  pounds 
balances  16  Ib.,  p  pounds  (one  half  of  the  weight  in  the 
left  pan)  must  balance  8  Ib.   (one  half  of  the  weight"  in 


4  GENERAL  MATHEMATICS 

the  right  pan).    By  the  use  of  the  equation  the  discussion 
may  be  written  in  the  following  brief  form: 

f  This  is  a  translation  of  the  tirst 
"  J  \      sentence  of  the  problem. 

Subtracting  1  from  "1 — — - 

each  member,      /    ^ 

Dividing  each  member  of  the  equation  by  2, 
p=8. 

This  problem  illustrates  the  principle  that  if  both  mem- 
bers of  an  equation  are  divided  by  the  name  number  (exclud- 
ing division  by  zero,  to  be  explained  later),  the  quotients  are 
equal;  that  is,  another  equation  is  obtained.  [Division  Law~\ 

EXERCISES 

Find  the  value  of  the  unknown  numbers,  doing  all  you 
can  orally: 

1.  2  x  +  3  =  9.  12.  5  r  +  ^2-  =  13|. 

2.  3x  +  4  =  16.  13.  14 k  +  7  =  79. 

3.  2  a  +  5  =  17.  14.  3  e  +  4j  =  9. 

4.  3*4-7  =  28.  15.  15x4-0.5  =  26. 

5.  5  r  4-  7  =  62.  16.  11  m  +  J  =  ^ 

6.  9s  4- 21  =  93.  17.  1.3y  +  3  =  16. 

7.  2y  +  l=S.  18.  11  y  =  33. 

8.  5  y  4-  3  =  15.  19.  1.1  x  =  121. 

9.  4x4-3.2=15.2.  20.  2.3x  +  4  =  50. 

10.  6^4-4  =  49.  21.  6.3  z  4-  2.4  =  15. 

11.  9e  + 8  =  116.  22.  5.3x4-0.34  =  2.99. 

5.  Addition  Law.  In  Fig.  3  the  apparatus  is  so  arranged 
that  the  2-pound  weight  attached  to  the  string  which 
passes  over  the  pulley  pulls  upward  on  the  bar  at  B.  This 


THE  EQUATION 


-Pulley 


arrangement  makes  the  problem  different  from  the  two 
which  we  have  considered.  If  there  were  no  pulley  attach- 
ment, the  weight  pulling  downward  in  the  left  pan  would 
be  5  a;  pounds.  Since  there  is  a  lifting  force  of  2  Ib.  at  B, 
the  downward  pulling  force  in  the  left  pan  is  2  Ib.  less 
than  5  x  pounds,  or  5  x  —  2  pounds  ;  this  balances  the  18  Ib. 
in  the  right  pan.  Hence  the  equation  which  describes  the 
situation  in  Fig.  3 
is  5z-2  =  18.  li- 
the string  be  cut  so 
as  to  remove  the  up- 
ward pull  of  2  Ib., 
then  a  2-pound 
weight  must  be 
added  to  the  right 
pan  to  keep  the 
scales  balanced,  for 
removing  the  up- 
ward pull  of  2  Ib. 
gave  us  a  down- 
ward pull  in  the  left  pan  of  5  a?  pounds.  This  is  2  Ib.  more 
than  we  had  with  the  pulley  attached,  hence  the  necessity 
of  adding  2  Ib.  to  the  right  pan. 

By  the  use  of  the  equation  the  preceding  discussion  may 
take  the  following  brief  form: 

.   .  _  o  _  i  o  /  This     expresses    the 
o  \.     original  conditions. 

£    —       M 

Adding  2  to  both  members, 


FIG.  3.   IN  THIS  CASE  THE  SCALES  ILLUSTRATE 
THE  ADDITION  LAW 


5  x 


=  20 


Dividing  both  members  by  5, 


This  problem  illustrates  the  principle  that  if  the  same 
number  is  added  to  both  members  of  an  equation,  the  sums  are 
equal ;  that  is,  another  equation  is  obtained.  [Addition  Law\ 


6  GENERAL  MATHEMATICS 

EXERCISES 

Find  the  value  of  the  unknown  number  in  each  problem, 
doing  all  you  can  orally  : 

1.  x  -5  =  10.  12.  12m  -35  =  41. 

2.  2 a; -15  =  13.  13.  9c-  3.2  =  14.8. 

3.  3  a; -12  =  13.  14.  7  *  -  2  =  5.7. 

4.  3.r-8  =  17.  15.  14  A- -  5  =  21. 

5.  12  y  -  4  =  46.            .  16.  2  y  -  3.1  =  3.2. 

6.  4 1  -  16  =  16.  17.  0.5  x  -  3  =  4.5. 

7.  19,--4i  =  14f  18.  2  x -1=61 

8.  lly-9  =  79.  19.  3cc-9l  =  17.r>. 

9.  56-  -  0.1  =  0.9.  20.  9.7-  -7.5  =  73.5. 

10.  4y-f  =  7j.  21.   1.5 a; -3  =  1.5. 

11.  7*- 4  =  26.  22.  1.6  x- 1.7  =  1.5. 

6.  Solving  an  equation ;  check ;  root.  The  process  of 
finding  the  value  of  the  unknown  number  in  an  equation 
is  called  solving  the  equation.  To  illustrate : 

Let  y  +  3  =  8  be  the  equation. 

"3  =  3 
Then  y  —  5,  and  the  equation  is  said  to  be  solved. 

To  test-,  or  check,  the  correctness  of  the  result  replace 
the  unknown  number  in  the  original  equation  by  5,  ob- 
taining 5  +  3  =  8.  Since  both  members  of  the  equation 
reduce  to  the  same  number,  the  result  y  =  5  is  correct. 

When  a  number  is  put  in  place  of  a  literal  number  it 
is  said  to  be  substituted  .for  the  literal  number. 

When  both  sides  of  an  equation  reduce  to  the  same 
number  for  certain  values  of  the  unknown  number,  the 
equation  is  said  to  be  satisfied.  Thus,  3  satisfies  the  equa- 
tion y  +  2  =  5. 


THE  EQUATION  7 

A  number  that  satisfies  an  equation  is  a  root  of  the 
equation. 

Thus,  5  is  a  root  of  the  equation  z  +  3  =  8. 

HISTORICAL  NOTE.  The  word  "root  "  first  appears  in  the  algebra 
of  Mohammed  ibn  Musa  Abu  Jafar  Al-Khwarizmi  (about  A.D.  830). 
The  root  of  an  equation  (like  the  root  of  a  plant)  is  hidden  until 
found.  See  Ball's  "A  Short  History  of  Mathematics,"  p.  163. 

EXERCISES 
•    Solve  the  following  equations  and  check  the  results : 

1.  5y+3=18.  5.  26  +  2.7-1.3=11.4. 

2.  7z-4=17.  6.  7x-3x  +  3.1  =  7.1. 

3.  2 a; -1.3  =  2.7.  7.  5* +14 -9  =15. 

4.  3a  +  4.5=7.5.  8.  7m  — 3f  =  3j. 

7.  Terms;  monomial;  order  of  terms.    The  parts  of  an 
expression  separated  by  plus  (+)  and  minus  (— )  signs  are 
called  the  terms  of  a  number.    Thus,  2  a  and  3  b  are  the 
terms  of  the  number  2  a-\-  3  b.  A  one-term  number  is  called 
a  monomial. 

EXERCISES 
1.8-7+2  =  ?  4.  8x—7x  +  2x  =  ? 

2.8  +  2-7=?  5.  8x  +  2x-7x  =  ? 

3.2  +  8-7=?  6.  2z  +  8z  —  lx  =  ? 

These  problems  illustrate  the  principle  (to  be  discussed 
more  fully  later)  that  the  value  of  an  expression  is  un- 
changed if  the  order  of  its  terms  is  changed,  provided 
each  term  carries  with  it  the  sign  at  its  left.  If  no  sign 
is  expressed  at  the  left  of  the  first  term  of  an  expression, 
the  plus  sign  is  understood. 

8.  Similar  and  dissimilar  terms.    Terms  which  have  a 
common  literal  factor,  as  2  r,  3  x,  and  5  x,  are  similar  terms. 
Their  sum  is  a  one-term  expression ;  namely,  10  a;.    When 


8  GENERAL  MATHEMATICS 

terms  do  not  have  a  common  literal  factor,  as  2  x  and 
3  y,  they  are  called  dissimilar  terms.  Algebraic  expressions 
are  simplified  by  combining  similar  terms.  Combining 
similar  terms  in  either  the  right  or  the  left  member  of  an 
equation  gives  us  the  same  equation  in  simpler  form. 

HISTORICAL  XOTE.  The  word  "algebra"  first  appears  about 
A.D.  830  in  an  Arabian  work  called  "  Al-jebr  wa'1-mukabala,"  written 
by  Al-Khwariznii.  "Al-jebr,"  from  which  "algebra"  is  derived,  may 
be  translated  by  the  restoration  and  refers  to  the  fact  that  the  same 
number  may  be  added  to  or  subtracted  from  both  sides  of  the 
equation;  "al-mukabala"  means  the  process  of  comparison,  and  some 
writers  say  it  was  used  in  connection  with  the  combination  of 
similar  terms  into  one  term. 

The  mathematical  interest  of  the  Arabs  ran  high.  In  the  seventh 
century  religious  enthusiasm  had  banded  these  nomadic  tribes 
into  a  conquering,  nourishing  nation.  Enormous  fortunes  demanded 
mathematical  manipulation.  Cantor  cites  a  rumor  of  a  merchant 
whose  annual  income  was  about  seven  million  dollars  and  a  Christian 
doctor  of  medicine  whose  annual  income  was  about  fifty  thousand. 
These  fortunes  gave  them  the  necessary  leisure  time  for  culture 
and  learning.  Among  the  many  books  translated  was  the  Greek 
geometry,  Euclid's  "  Elements."  See  Ball's  K  A  Short  History  of 
Mathematics,"  p.  162,  and  Miller's  "Historical  Introduction  to 
Mathematical  Literature,"  p.  83. 

EXERCISES 

Solve  the  following  equations  and  check  the  results  : 

1.  2./-  -7  =x  +  3. 

Subtract  x  from  both  members  and  proceed  as  usual. 

2.  3x  +  2  =  x  +  S. 

3.  5ix  —  3x  +  2-x  —  2  =  2x  + 

4.  16y-8y  +  3y-2  =  5y 

5.  20-4.r  =  38  -10 a-. 

6.  5.T  -j-3  — 05  =  .r  +18. 


THE  EQUATION  9 

• 

7.  7  >•  +18  +  3  r  =  32  +  2  r  -  2. 

8.  18  +  6  s  +  30  +  6  s  =  4  s  +  8  +  12  +  3  *  +  3  +  s  +  29. 

9.  25  y  +  20  -  7  y  -  5  =  56  -  5  //  +  5. 

^J 

9.  Multiplication  Law.    Solve  for  x:  -  =  5. 

A 

This  problem  offers  a  new  principle.  If  we  translate  it 
into  an  English  sentence,  it  reads  as  follows  :  One  half  of 
what  unknown  number  equals  5  ?  If  one  half  of  a  num- 
ber equals  5,  all  of  the  number,  or  twice  as  much,  equals 
2  times  5,  or  10.  The  problem  may  be  solved  as  follows: 


Multiplying  both  members  by  2, 

2  x      =  2  -x  5. 


x  =  10. 

The  principle  involved  here  may  be  further  illustrated 
by  the  scales,  for  if  an  object  in  one  pan  of  a  scales  will 
balance  a  5-pound  weight  in  the  other,  it  will  be  readily 
seen  that  three  objects  of  the  same  kind  would  need  15  Ib. 
to  balance  them.  This  may  be  expressed  in  equation  form 
as  feitews  :  _  = 

X  —  U. 

Mujtiplyijig  both  members  by  3, 

3  x  =  15. 

v. 

From  the  preceding'  discussion  it  is  evident  that  if 
both  members  of  an  equation  are  multiplied  by  the  same 
number,  the  products  are  equal:  that  is,  another  equation 
is  obtained.  [Multiplication  Law\ 


10  GENERAL  MATHEMATICS 

This  multiplication  principle  is  convenient  when  an 
equation  contains  fractions.  It  enables  us  to  obtain  a 
second  equation  containing  no  fraction  but  containing  the 
same  unknown  number.  To  illustrate  this: 

Let  \  .-•  =  7. 

Multiplying  both  members  by  3, 

3  x  i-  x  =  3  x  7. 

Reducing  to  simplest  form,          x  =  21. 

ORAL  EXERCISES 

Find  the  value  of  the  unknown  number  in  each  of  the 
following  equations  : 

• 

tn  />  &  IT 


The  preceding  list  of  problems  shows  that  it  is  desirable 
to  multiply  both  members  of  the  equation  by  some  number 
that  will  give  us  a  new  equation  without  fractions.  The  same 
principle  holds  when  the  equation  contains  two  or  more 
fractions  whose  denominators  are  different,  as  is  illustrated 
by  the  following  problem : 

Find  x  if     7-|  =  2. 
•4      5 

Solution.  -  -  -  =  2. 

4      5 

90  r        90  r 

Multiplying  by  20,  —  =  40. 

Simplifying,  5  x  —  4  x  —  40  ; 

whence  x  =  40. 


THE  EQUATION  11 

The  fact  that  4  and  5  will  divide  integrally  (a  whole 
number  of  times)  into  the  numerators  gives  us  a  new  equa- 
tion without  fractions.  Obviously  there  are  an  unlimited 
number  of  numbers  (for  example,  40,  60,  80,  etc.)  which 
we  could  have  used,  but  it  was  advantageous  to  use  the 
smallest  number  in  which  4  and  5  are  contained  integrally  ; 
namely,  the  least  common  multiple  of  4  and  5,  which  is 
20.  The  object  of  this  multiplication  is  to  obtain  an  equa- 
tion in  which  the  value  of  the  unknown  number  is  more 
easily  found  than  in  the  preceding  one.  This  discussion 
may  be  summarized  by  the  following  rule: 

If  the  given  equation  contains  fractions,  multiply  every  term 
in  both  members  by  the  least  common  multiple  {L.C.M.}  of 
the  denominators  in  order  to  obtain  a  new  equation  which  does 
not  contain  fractions. 

EXERCISES 

Find  the  value  of  the  unknown  number  in  each  problem, 
and  check  : 


2    ^  +  ^-9  a    2x   ,x        x     ,    l 

<S«     „     i      .    —   <?•     •  H.    -  -f-  —  =  -  +  —  • 

54  96      18      3 


4.  it  +  J  y  =  6.  '  K 

11    ^ 

5-  y  -  I  y  =  7.  6 


12  GENERAL  MATHEMATICS 

10.  Definition  of  the  equation;  properties  of  the  equation. 

The  foregoing  problems  were  used  to  show  that  an  equa- 
tion iif  a  statement  that  two  numbers  are  equal.  It  indicates 
that  two  expressions  stand  for  the  same  number.  It  may 
be  regarded  as  an  expression  of  balance  of  values  be- 
tween the  numbers  on  the  two  sides  of  the  equality  sign. 
Some  unknown  number  which  enters  into  the  discussion 
of  the  problem  is  represented  by  a  letter.  An  equation 
is  written  which  enables  one  to  find  the  value  of  that 
unknown  number. 

An  equation  is  like  a  balance  in  that  the  balance  of 
value  is  not  disturbed  so  long  as  like  changes  are  made 
on  both  sides.  Thus,  in  equations  we  may  add  the  same 
number  to  both  aides,  or  subtract  the  same  number  from 
both  sides,  or  we  may  multiply  or  divide  both  sides  by  the 
same  number  (except  division  by  zero);  the  equality  is 
maintained  during  all  these  changes. 

On  the  other  hand,  the  equality  is  destroyed  if  more  is 
added  to  or  subtracted  from  one  side  than  the  other  or  if 
one  side  is  multiplied  or  divided  by  a  larger  number  than 
is  the  other  side. 

11.  Translation  of  an  equation.   The  list  of  problems  in 
the  preceding  exercises  may  appear  abstract  in  the  sense 
that  the  equations  do  not  appear  to  be  connected  in  any 
way  either  with  a  concrete  situation  of  a  verbal  problem, 
or  with  our  past  experience.    However,  such  a  list  need 
not  be  regarded  as  meaningless.    Just  as  an  English  sen- 
tence which  expresses  a  number  relation  may  be  written 
in  the  "  shorthand  "  form  of  an  equation,  so,  conversely, 
an  equation  may  be  translated  into  a  problem ;    for  ex- 
ample, the  equation  3x  +  5  =  2x  +  2Q  may  be  interpreted 
as  follows :   Find  a  number  such  that  3  times  the  num- 
ber plus    5    equals   2  times  the   number   plus   20.     The 


THE  EQUATION  13 

equation  .c— 3  =  5  may  be  regarded  as  raising  the  question 
What  number  diminished  by  3 '  equals  5  ?  Or,  again, 
21  x  +  x  +  2±x  +  x  =  140  may  be  considered  as  the  trans- 
lation of  the  following  problem:  What  is  the  altitude  of 
a  rectangle  whose  base  is  21  times  as  long  as  the  altitude 
and  whose  perimeter  is  140  ft.  ? 

EXERCISES 

State  each  of  the  following  in  the  form  of  a  question  or  a 
verbal  problem  : 

1.  05-6  =  3.  7.  7r-2  =  6r  +  8. 

2.  2a- -1  =  10.  8.  5.2 x-  3  =  4.1  x+ 1.4. 

3.  9 k  -10  =  87.  9.  3*  =12. 

4.  ly  +  8  =  112.  10.  4*  =16. 

5.  7s- 3  =  81.  11.  2x  +  3x  +  4x  =  18. 

6.  3x  +  2  =  2x  +  :;.  12.  c  =  2T2  rl. 

12.  Drill  in  the  "shorthand"  of  algebra.  The  following 
exercises  give  practice  in  translating  number  expressions 
and  relations  from  verbal  into  symbolical  language: 

1.  Consecutive  numbers  are  integral  (whole)  numbers  which 
follow  each  other  in  counting ;  thus,  17  and  18,  45  and  46,  are 
examples  of  consecutive  numbers.    Begin  at  s  -f  5  and  count 
forward.  Begin  at  x  +  3  and  count  backward.    Give  four  con- 
secutive integers  beginning  with  18  ;  ending  with  18  ;  beginning 
with  x;  ending  with  x.    Give  two  consecutive  even  integers 
beginning  with  2  x.    Give  two  consecutive  even  integers  end- 
ing with  2c. 

2.  The  present  age,  in  years,  of  a  person  is  denoted  by  x. 
Indicate  in  symbols  the  following :  (a)  the  person's  age  fourteen 
years  ago ;  (b)  his  age  fourteen  years  hence ;  (c)  his  age  when 
twice  as  old  as  now ;  (d)  60  decreased  by  his  age ;  (e)  his  age 
decreased  by  60 ;  (f)  his  age  increased  by  one  half  his  age. 


14  GENERAL  MATHEMATICS 

3.  A  boy  has  a  marbles  and  buys  b  more.    How  many  has 
he  ?    What  is  the  sum  of  'a  and  b  ? 

4.  A  boy  having  b  marbles  loses  c  marbles.    How  many 
has  he? 

5.  (a)  The  home  team  made  8  points  in  a  basket-ball  game 
and  the  visiting  team  made  3  points.    By  how  many  points  did 
the  home  team  win  ?  (b)  If  the  home  team  scored  h  points 
to  the  visitors'  n  points,  by  how  many  points  did  the  home 
team  win  ?    (c)  Substitute  numbers  for  h  in  the  last  question 
that  will  show  the  defeat  of  the  home  team,   (d)  If  h  =  5,  what 
must  be  the  value  of  n  when  the  game  is  a  tie  ? 

6.  What  is  the  5th  part  of  n  ?    f  of  y  ?    f  of  t  ? 

7.  Two  numbers  differ  by  7.    The  smaller  is  s.    Express 
the  larger  number. 

8.  Divide  100  into  two  parts  so  that  one  part  is  «. 

9.  Divide  a  into  two  parts  so  that  one  part  is  5. 

10.  The  difference  between  two  numbers  is  d  and  the  larger 
one  is  I.    Express  the  smaller  one. 

11.  What  number  divided  by  3  will  give  the  quotient  a? 

12.  A  man's  house,  worth  h  dollars,  was  destroyed  by  fire. 
He  received  i  dollars  insurance.    What  was  his  total  loss  ? 

13.  A  is  x  years  old  and  B  lacks  5  yr.  of  being  three  times 
as  old.    Express  B's  age. 

14.  A  has  m  ties  and  B  has  n  ties.    If  A  sells  B  5  ties,  how 
many  will  each  then  have  ? 

15.  A  man  has  d  dollars  and  spends  c  cents.    How  many 
cents  does  he  have  left  ? 

16.  A  room  is  I  feet  long  and  w  feet  wide.    How  many  feet 
of  border  does  such  a  room  require  ? 

17.  The  length  of  a  rectangle  exceeds  its  width  by  c  feet. 
It  is  w  feet  wide,    (a)  State  the  length  of  each  side,    (b)  Find 
the  distance  around  the  rectangle. 


THE  EQUATION  15 

18.  What  is  the  cost  of  7  pencils  at  c  cents  each  ? 

19.  What  is  the  cost  of  1  sheet  of  paper  if  b  sheets  can  be 
bought  for  100? 

20.  It  takes 'two  boys  5  da.  to  make  an  automobile  trip. 
What  part  can  they  do  in  1  da.  ?     If  it  takes  them  d  days, 
what  part  of  the  trip  do  they  travel  in  1  da.  ? 

21.  If  a  man  drives  a  car  at  the  rate  of  31  mi.  per  hour, 
how  far  can  he  drive  in  3  hr.  ?    in  5  hr.  ?    in  li  hours  ? 

22.  If  a  man  drives  n  miles  in  3  hr.,  how  many  miles  does 
he  go  per  hour  ? 

23.  A  tank  is  filled  by  a  pipe  in  m  minutes.    How  much  of 
the  tank  is  filled  in  1  min.  ? 

24.  The  numerator  of  a  fraction  exceeds  the  denominator 
by   3.     (a)  Write   the   numerator,     (b)  Write   the    fraction, 
(c)  Head  the  fraction. 

25.  A  pair  of  gloves  costs  d  dollars.   What  is  the  cost  if  the 
price  is  raised  70?  if  lowered  70? 

26.  Write  the  sum  of  x  and  17 ;  of  17  and  x.    Write  the 
difference  of  x  and  17;    of  17  and  x. 

27.  A  class  president  was  elected  by  a  majority  of  7  votes. 
If  the  unsuccessful  candidate  received  k  votes,  how  many  votes 
were  cast  ? 

13.  Algebraic  solution.  Many  problems  may  be  solved 
by  either  arithmetic  or  the  use  of  the  equation.  When 
the  solution  of  a  problem  is  obtained  by  the  use  of  the 
equation, 'it  is  commonly  called  an  algebraic  solution. 

The  following  problems  illustrate  the  important  steps 
in  the  algebraic  solution  of  a  problem.  By  way  of  contrast 
an  arithmetic  solution  is  given  for  the  first  problem. 

1.  Divide  a  pole  20  ft.  long  into  two  parts  so  that  one  part 
shall  be  four  times  as  long  as  the  other. 


16  GEXKKAL  .MATHEMATICS 

ARITHMETICAL  SOLUTION 
The  shorter  part  is  a  certain  length. 
The  longer  part  is  four  times  this  length. 
The  whole  pole  is  then  five  times  as  long  as  the  shorter  part. 
The  pole  is  20  ft.  long. 
The  shorter  part  is  \  of  20  ft.,  or  4  ft. 
The  longer  part  is  4  x  4  ft.,  or  16  ft. 
Hence  the  parts  are  4  ft.  and  16  ft.  long  respectively. 

ALGEBRAIC  SOLUTION 

Let  n  =  number  of  feet  in  the  shorter  part. 

Then  4  n  =  number  of  feet  in  the  longer  part, 

and  n  +  4  n,  or  5  n  —  length  of  the  pole. 

Then  .">  n  =  '20. 

n  =  4. 
4  n  =  10. 

Hence  the  parts  are  4  ft.  and  1(J  it.  long  respectively. 
/ 
2.  A  rectangular  garden  is  three  times  as  long  as  it  is  wide. 

It  takes  80  yd.  of  fence  to  inclose  it.     Find  the  width  and 

length. 

ALGEBRAIC  SOLUTION 

Let  x  =  number  of  feet  in  the  width. 

Then  .3  x  =  number  of  feet  in  the  length, 

and  ./•  +  :>  x  +  x  +  3  x  =  distance  around  the  garden. 

Then  8  x  =  80. 

x  =  10. 
3  x  =  30. 
Hence  the  width  is  10  yd.  and  the  length  is  30  yd. 

14.  The  important  steps  in  the  algebraic  solution  of  verbal 
(or  story)  problems.  Before  proceeding  to  the  solution  of 
more  difficult  problems  it  is  important  that  we  organize  the 
steps  that  are  involved.  The  preceding  list  of  problems  illus- 
trates the  following  method  for  solving  a  verbal  problem : 

(a)  In  every  problem  certain  facts  are  given  as  known 
and  one  or  more  as  unknown  and  to  be  determined.  Read 
the  problem  so  as  to  get  these  facts  clearly  in  mind. 


THE  EQUATION  17 

(b)  In  solving  the  problem  denote  one  of  the  unknown 
numbers  by  some  symbol,  as  y. 

(c)  Then  express  all  the  given  facts  in  algebraic  lan- 
guage, using  the  number  y  as  if  it,  too,  were  known. 

(d)  Find  two  different  expressions  which  denote  the 
same   number  and   equate    them.     (Join   by  the   sign   of 
equality  (=).) 

(e)  Solve  the  equation  for  the  value  of  the  unknown 
number. 

(f)  Check  the  result  by  re-reading  the  problem,  substi- 
tuting the  result  in  the  conditions  of.  the  problem  to  see 
if  these  conditions  are  satisfied.    Note  that  it  is  not  suffi- 
cient to  check  the  equation,  for  you  may  have  written  the 
wrong  equation  to  represent  the  conditions  of  the  problem. 

EXERCISES 

With  the  preceding  outline  of  method  in  mind,  solve  the 
following  problems,  and  check : 

PROBLEMS  INVOLVING  NUMBER  RELATIONS 

1.  If  six  times  a  number  is  decreased  by  4,  the  result  is  26. 
Find  the  number. 

2.  If  four  fifths  of  a  number  is  decreased  by  6,  the  result 
is  10.    Find  the  number. 

3.  The  sum  of  three  numbers  is  12Q.    The  second  is  five 
times  the  first,  and  the  third  is  nine  times  the  first.    Find 
the  numbers. 

4.  The  sum  of  three  numbers  is  360.    The  second  is  four- 
teen times  the  first,  and  the  third  is  the  sum  of  the  other  two. 
Find  the  numbers. 

5.  Seven  times  a  number  increased  by  one  third  of  itself 
equals  44.    Find  the  number. 


18  GENERAL  MATHEMATICS 

6.  The  following  puzzle  was  proposed  to  a  boy :  "  Think 
of  a  number,  multiply  it  by  4,  add  12,  subtract  6,  and  divide 
by  2."    The  boy  gave  his  final  result  as  13.    What  was  his 
original  number  ? 

7.  The  sum  of  one  half,  one  third,  and  one  fourth  of  a 
number  is  52.    What  is  the  number  ? 

CONSECUTIVE-NUMBER    PROBLEMS 

8.  Find  two  consecutive  numbers  whose  sum  is  223. 

9.  Find  three  consecutive  numbers  whose  sum  is  180. 

10.  Find  two  consecutive  odd  numbers  whose  sum  is  204. 

* 

11.  Find  three  consecutive  even  numbers  whose  sum  is  156. 

12.  It  is  required  to  divide  a  board  70  in.  long  into  five  parts 
such  that  the  four  longer  parts  shall  be  1",  2",  3",  and  4" 
longer  respectively  than  the  shortest  part.    Find  the  lengths 
of  the  different  parts. 

13.  A  boy  in  a  manual-training  school  is  making  a  bookcase. 
The  distance  from  the  top  board  to  the  bottom  is  4  ft.  7  in., 
inside  measure.    He  wishes  to  put  in  three  shelves,  each  1  in. 
thick,  so  that  the  four  book  spaces  will  diminish  successively 
by  2  in.  from  the  bottom  to  the  top.    Find  the  spaces.    . 

PROBLEMS  INVOLVING  GEOMETRIC  RELATIONS 

14.  The  length  of  a  field  is  three  times  its  width,  and  the 
distance  around  the  field  is  200  rd.    If  the  field  is  rectangular, 
what  are  the  dimensions  ? 

15.  A  room  is  15  ft.  long,  14  ft.  wide,  and  the  walls  contain 
464  sq.  ft.    Find  the  height  of  the  room. 

16.  The  perimeter  of  (distance  around)  a  square  equals  64  ft. 
Find  a  side. 

NOTE.  Such  geometric  terms  as  "triangle,"  "rectangle,"  "square," 
etc.,  as  occur  in  this  list  of  problems  (14-24),  are  familiar  from  arith- 
metic. However,  they  will  later  be  defined  more  closely  to  meet 
other  needs. 


THE  EQUATION  19 

17.  Find  the  sides  of  a  triangle  if  the  second  side  is  3  ft. 
longer  than  the  first,  the  third  side  5  ft.  longer  than  the  first, 
and  the  perimeter  is  29  ft. 

18.  Find  the  side  of  an  equilateral  (all  sides  equal)  triangle 
if  the  perimeter  is  21f  ft. 

19.  The  perimeter  of  an  equilateral  pentagon  (5-sided  figure) 
is  145  in.    Find  a  side. 

20.  The  perimeter  of  an  equilateral  hexagon  (6-sided  figure) 
is  192  ft.    Find  a  side. 

21.  Find  the  side  of  an  equilateral  decagon  (10-sided  figure) 
if  its  perimeter  is  173  in. 

22.  What  is  the  side  of  an  equilateral  dodecagon  (12-sided 
figure)  if  its  perimeter  is  288  in.  ? 

23.  A  line  60  in.  long  is  divided  into  two  parts.    Twice  the 
larger  part  exceeds  five  times  the  smaller  part  by  15  in.    How 
many  inches  are  in  each  part  ? 

24.  The  perimeter  of  a  quadrilateral  A  BCD  (4-sided  figure) 
is  34  in.    The  side  CD  is  twice  as  long,  as  the  side  AB;  the 
side  AD  is  three  times  as  long  as  CD;  the  side  BC  equals  the 
sum  of  the  sides  AD  and  CD.    Find  the  length  of  each  side. 

MISCELLANEOUS  PROBLEMS 

25.  Divide  $48,000  among  A,  B,  and  C  so  that  A's  share 
may  be  three  times  that  of  B,  and  C  may  have  one  half  of 
what  A  and  B  have  together. 

26.  The  perimeter  of  a  rectangle  is  132  in.    The  base  is 
double  the  altitude.    Find  the  dimensions  of  the  rectangle. 

27.  A  and  B  own  a  house  worth  $16,100,  and  A  has  invested 
twice  as  much  capital  as  B.    How  much  has  each  invested  ? 

28.  A  regulation  football  field  is  56|^  yd.  longer  than  it  is 
wide,  and  .the  sum  of  its  length  and  width  is  163^  yd.    Find 
its  dimensions. 


20  GENERAL  MATHEMATICS 

29.  A  man  has  four  times  as  many  chickens  as  his  neighbor. 
After  selling  14,  he  has  3^  times  as  many.    How  many  had 
each  before  the  sale  ? 

30.  In  electing  a  president  of  the  athletic  board  a  certain 
high  school  cast  1019  votes  for  three  candidates.    The  first 
received  143  more  than  the  third,  and  the  second  49  more 
than  the  first.    How  many  votes  did  each  get  ? 

31.  A  boy  has  $5.20  and  his  brother  has  $32.50.    The  first 
saves  200  each  day  and  the  second  spends  100  each  day.    In 
how  many  days  will  they  have  the  same  amount  ? 

32.  One  man  has  seven  times  as  many  acres  as  another. 
After  the  first  sold  9  A.  to  the  second,  he  had  36  A.  more 
than  the  second  then  had.    How  many  did  each  have  before 
the  sale  ? 

33.  To  find  the  weight  of  a  golf  ball  a  man  puts  20  golf 
balls  into  the  left  scale  pan  of  a  balance  and  a  2-pound  weight 
into  the  right;    he  finds  that  too  much, f but  the  balance  is 
restored  if  he  puts  2.  oz.  into  the  left  scale  pan.    What  was 
the  weight  of  a  golf  ball  ? 

34.  The  number  of  representatives  and  senators  together 
in  the  United  States  Congress  is  531.    The  number  of  repre- 
sentatives is  51  more  than  four  times  the  number  of  senators. 
Find  the  number  of  each. 

35.  A  boy,  an  apprentice,  and  a  master  workman  have  the 
understanding  that  the  apprentice  shall  receive  twice  as  much 
as  the  boy,  and  the  master  workman  four  times  as  much  as 
the  boy.    How  much  does  each  receive  if  the  total  amount 
received  for  a  piece  of  work  is  $105  ? 

36.  A  father  leaves  $13.000  to  be  divided  among  his  three 
children,  so  that  the  eldest  child  receives  $2000  more  than 
the  second,  and  twice  as  much  as  the  third.    What  is  the 
share  of  pa oh  ? 


THE  EQUATION  21 

37.  A  fence  5  ft.  high  is  made  out  of  6-inch  boards  running 
lengthwise.    The   number  of   boards  necessary  to   build  the 
fence  up  to  the  required  height  is  5  and  they  are  so  placed 
as  to  leave  open  spaces  between  them.    If  each  of  these  open 
spaces,  counting  from  the  bottom  upwards,  is  half  of  the  one 
next  above  it,  what  must  be  the  distances  between  the  boards ; 
that  is,  what  will  be  the  width  of  each  of  the  open  spaces  ? 

38.  I  paid  $8  for  an  advertisement  of  8  lines,  as  follows : 
240  a  line  for  the  first  insertion,  100  a  line  for  each  of  the 
next   five   insertions,  and   20   a   line    after   that.     Firid   the 
number  of  insertions. 

39.  The  annual  income  of  a  family  is  divided  as  follows : 
One  tenth  is  used  for  clothing,  one  third  for  groceries,  and  one 
fifth  for  rent.    This  leaves  $660  for  other  expenses  and  for 
the  savings  account.    How  much  is  the  income  ? 

15.  Axioms.  Thus  far  we  have  used  the  four  following 
laws  in  solving  equations: 

I.  If  the  same  number  be   added   to   equal  numbers,   the 
sums  are  equal.    [Addition  Law] 

II.  If  the  same  number  be  subtracted  from  equal  numbers, 
the  remainders  are  equal.    [Subtraction  Law] 

III.  If  equal  numbers  be  multiplied  by  the  same  number, 
the  products  are  equal.    [Multiplication  Law] 

IV.  If  equal   numbers  be   divided   by   the   same  number 
(excluding  division  by  zero),  the  quotients  are  equal.    [Division 
Law] 

Statements  like  the  four  laws  above,  when  assumed  to 
be  true,  are  called  axioms.  Usually  axioms  are  statements 
so  simple  that  they  seem  evident.  A  simple  illustration 
is  sufficient  to  make  clear  the  validity  of  the  axiom.  For 
example,  if  two  boys  have  the  same  number  of  marbles 
and  3  more  are  given  to  each,  then  our  experience  tells 


22  GENERAL  MATHEMATICS 

us  that  again  one  boy  would  have  just  as  many  as  the 
other.  This  illustrates  the  validity  of  the  addition  axiom. 
Hereafter  the  preceding  laws  will  be  called  Axioms  I,  II, 
III,  and  IV  respectively. 

16.  Axiom  V.  In  this  chapter  we  have  also  made  fre- 
quent use  of  another  axiom.  In  solving  a  verbal  problem 
we  obtained  the  necessary  equation  by  finding  two  expres- 
sions which  denoted  the  same  number  and  then  we  equated 
these  two  expressions.  This  step  implies  the  following 
axiom : 

If  two  numbers  are  equal  to  the  same  number  (or  to  equal 
numbers'),  the  numbers  are  equal.  [Equality  Axiom] 

Illustrate  the  truth  of  Axiom  V  by  some  familiar 
experience. 

The  following  exercises  test  and  review  the  axioms. 

EXERCISES 

Solve  the  following  equations,  and  check.  Be  able  to  state 
at  every  step  in  the  solution  the  axiom  used. 

1.  12  #-15  =  30. 

Solution.  12 1  -  15    =30 

15    =  15  (Axiom  I) 

Adding  15  to  both  members,  12  t  =  45 

Dividing  both  sides  by  12,  /  =  f  f,  or  3£.    (Axiom  IV) 

y     y     1 

2.  -  +  -  =  -• 

24       3 

Solution.  Multiplying  both  sides  of  the  equation  by  the  least 
common  multiple  of  the  denominators,  that  is,  by  12, 

12  y  ,  12  y      12 

-2^  +  -^:=  y  (Axiom  III) 

Then  6  y  +  3  y  =  4. 


THE  EQUATION  23 

By  reducing  the  fractions  of  the  first  equation  to  lowest  terms 
we  obtain  the  second  equation,  which  does  not  contain  fractions. 

Combining  similar  terms,      9  y  —  4. 
Dividing  both  sides  of  the  equation  by  9, 

y  •=•  f.  (Axiom  IV) 

3.  12a-+13  =  73.  8.  17s-  3s  +  16s  =105. 

4.  18r-12r  =  33.  9.  17x  +  3x  -  9x  =  88. 

5.  21^+15=120.  10.  16m  +  2m  —  13m  =  22%. 

6.  28* -9  =  251.  11.  202y-152//  +  6?/  =  280. 

7.  20y+ 2y-18y  =  22.  12.  3.4 x  — 1.2 x +  4.8 x  =  70. 

13.  3.5  y  +  7.6  ?/-  8.6y=15. 

14.  5.8  m  —  3.9  m  + 12.6  m  =  58. 

15.  6  >•  -  3.5  >•  +  5.5  r  =  68. 

16.  3.41  x  +  0.59  x  - 1.77 x  =  22.3. 

17.  8  y  —  4.5  y  +  5.2  y  =  87. 

18.  2s  +7s  -  3s-  6  =  24. 

19.  lx  =  6.  27.  ^  +  |_|  =  36. 

20.  J*  =  2.  28.  15  =  3sc-3. 

21.  fa;  =  6.  Solution.    Adding   3    to  both 

members, 

22.  fa;  =  25.  18  =  3*. 

^       jj.  J)ividing  both  members  by  3, 

23-  3  +  2=1°-  6=x. 

Note   that  in   the   preceding 

24.  —  +  —  =  3.  problem   the   unknown    appears 

in  the  right  member. 

25.  ^  +  ^  =  8.  29.  17=2^-3. 

30. 

26.-^  =  6.  31. 


24  GENERAL  MATHEMATICS 

32.^  =  4.  33.^  =  5. 

x  b 

Solution.     Multiplying     both  _ .     16  _  « 

members   by  #,  3  x 

1Q  =  ix.  Multiply  both  members  by  3  x. 

Hence  4  =  x.  _  _      3 

<5o.   - —  =  1. 

4  £ 
Note  that   in   the   preceding 

problem  the  unknown  occurs  in  ,g     13          _  . 

the  denominator.  2  £ 

SUMMARY 

17.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  equation,  members  of  an  equation, 
equation  is  satisfied,  substituting,  check,  root  of  an  equa- 
tion, verbal  problem,  algebraic  solution  of  a  verbal  problem, 
term,  monomial,  literal  number,  similar  terms,  dissimilar 
terms,  order  of  terms,  and  axiom. 

18.  Axioms.    In  solving  equations  the  following  axioms 
are  used : 

I.  If  the  same  number  or  equal  numbers  be  added  to  equal 
numbers,  the  sums  are  equaL    [Addition  Axiom] 

II.  If  the  same  number  or  equal  numbers  be  subtracted  from 
equal  numbers,  the  remainders  are  equal.    [Subtraction  Axiom] 

III.  If  equal  tno/ihcrs  be  divided  by  equal  numbers  (exclud- 
ing division  by  zero),  the  quotients  are  equal.    [Division  Axiom] 

IV.  If  equal  numbers  be  multiplied  by  the  same  number  or 
equal  numbers,  the  products  are  equal.    [Multiplication  Axiom] 

V.  If  two  numbers  are  equal  to  the  same  number  or  to  equal 
numbers,  the  numbers  are  equal.    [Equality  Axiom] 

19.  If  an  equation  contains  fractions,  a  second  equation 
involving  the  same  unknown  may  be  obtained  by  multi- 
plying every  term  of  the  given  equation  by  the  L.C.M. 


THE  EQUATION  25 

of  the  denominators  and  then  reducing  all  fractions  to 
integers.  The  solution  of  this  equation  is  more  readily 
obtained  than,  that  of  the  given  equation. 

20.  The  equation  is  a  convenient  tool  for  solving 
problems.  In  solving  a  verbal  problem  algebraically  ob- 
serve the  following  steps: 

(a)  Use  a  letter  for  the  unknown  number  called  for  in 
the  problem.    Often  the  last  sentence  suggests  the  most 
convenient  choice. 

(b)  Express  the  given  facts  in  terms  of  the  unknown 
(provided,  of  course,  that  these  facts  are  not  definite  arith- 
metical numbers). 

(c)  Obtain  two  expressions  for  the  same  number  and 
equate  them.     (This  gives  us  the  equation.) 

(d)  Solve  the  equation.    Check  the  result  by  substitut- 
ing in  the  conditions  of  the  problem. 


CHAPTER  II 

LINEAR  MEASUREMENT.    THE  EQUATION  APPLIED  TO 

LENGTH1 

21.  Length,  the  important  characteristic  of  lines.  If  in 
drawing  an  object  we  lay  the  ruler  on  a  sheet  of  paper  and 
pass  a  sharpened  pencil  as  near  the  edge  as  possible,  thus 
obtaining  what  is  familiar  to  us  as  a  straight  line,  we  are  at 
once  concerned  with  the  length  of  a  part  of  the  straight  line 
drawn.  In  fact,  length  is  the  important  characteristic  of  a 
line.  In  an  exact  sense  a  line  has  length  only,  not  width  nor 

thickness.      Thus,     A ^ 

the  edge  of  a  table 

FIG.  4.  A  LINE  SEGMENT 
has    length    only ; 

the  thickness  and  width  of  a  crayon  line  are  neglected ;  the 
wide  chalk  marks  on  a  tennis  court  are  not  boundary 
lines,  but  are  made  wide  to  help  us  see  the  real  boundary 
lines,  which  are  the  outside  edges  of  the  chalk  marks. 

The  part  of  a  line  whose  length  we  wish  to  determine 
is  a  line  segment  or,  briefly,  a  segment,  as  AB  in  Fig.  4. 

A  line  segment  has  a  definite  beginning  point  and  a 
definite  ending  point.  The  word  "point"  is  used  to  mean 
merely  position,  not  length,  breadth,  nor  thickness.  The 
position  of  a  point  is  shown  by  a  short  cross  line ;  that  is, 
a  point  is  determined  by  two  intersecting  lines. 

1  The  pupil  should  now  provide  himself  with  a  ruler  one  edge  of  which 
is  graduated  to  inches  and  fractional  parts  of  an  inch  and  the  other  to 
units  of  the  metric  scale.  He  should  also  obtain  a  pair  of  compasses 
and  some  squared  paper  ruled  to  the  metric  scale. 

26 


LINEAR  MEASUREMENT  27 

EXERCISE 

Give  examples  of  line  segments  that  can  be  seen  in  the 
classroom. 

22.  Measurement  of  length.  In  Fig.  5  the  length  of  the 
line  segment  AB  is  to  be  determined.  One  edge  of  your 
ruler  is  graduated  (divided)  into  inches  and  fractions  of 
an  inch  as  is  shown  in  Fig.  5.  Place  the  division  marked 
zero  on  your  ruler  at  A,  with  the  edge  of  the  ruler  along 

A  B 

i f 


FIG.  5.    How  A  LINE  SEGMENT  MAT  BE  MEASURED 

the  segment  AB,  and  read  the  number  of  inches  in  the  line 
segment  AB;  that  is,  find  what  reading  on  the  ruler  is 
opposite  the  point  B. 

In  the  preceding  problem  we  compared  the  unknown 
length  of  the  line  segment  AB  with  the  well-established 
and  well-known  segment,  the  inch,  and  found  the  line 
segment  to  be  21  times  as  long  as  the  inch.  Hence  the 
length  of  the  line  segment  is  2^  in.  When  we  determine 
the  length  of  a  line  segment  we  are  measuring  the  line 
segment.  The  segment  with  which  we  compare  the  given 
line  is  called  a  unit  segment  or  a  unit  of  measurement.  Hence 
to  measure  a  line  segment  is  to  apply  a  standard  unit  seg- 
ment to  it  to  find  out  how  many  times  the  unit  segment  is 
contained  in  it. 

EXERCISES 

1.  Draw  a  line  segment  and  express  its  length  in  inches. 

2.  Measure  the  length  of  your  desk  in  inches. 

3.  Measure  the  width  of  your  desk  in  inches. 


28  (rEXEKAL   MATHEMATICS 

23.  Different  units  of  length.  The  most  familiar 
inents  for  measuring  are  the  foot  rule,  the  yardstick,  and 
several  kinds  of  tape  lines.  The  fractional  parts  of  the  unit 
are  read  by  means  of  a  graduated  scale  engraved  or  stamped 
on  the  standard  unit  used.  In  Fig.  6,  below,  is  shown  a 
part  of  a  ruler.  The  upper  edge  is  divided  into  inches 
and  fractional  parts  of  an  inch.  What  is  the  length  of  the 
smallest  line  segment  of  the  upper  edge  ? 

The  lower  edge  of  the  ruler  is  divided   into  units  of 
the  metric  (or  French)  scale.    This  system  is  based  on  the 


I '  i '  I '  I '  '  t '  i '  i '  '  i '  i '  i '  i '  M  i '  i '  M  1 1 1 1 1 '  M  M  i  f  m  1 1 1 1 '  1 1 1 

O    INCH  i  2  3 

O  1  2  3  4 

I  ..    CENTIMETERS   |         .  ,    .  ..| ....    I. 

u 

'  I-> 

iCm. 
FIG.  6.    PART  OF  A  RULER,  SHOWING  DIFFERENT  UNITS  OF  LENGTH 

decimal  system  and  is  now  very  generally  used  in  scien- 
tific work  in  all  countries.  The  standard  unit  is  called  the 
meter  (m.).  It  is  divided  into  1000  equal  parts  called  milli- 
meters (mm.).  In  the  figure  above,  the  smallest  division 
is  a  millimeter.  Ten  millimeters  make  a  centimeter  (cm.). 
In  the  figure,  AB  is  one  centimeter  in  length,  and  you  will 
note  that  this  is  about  two  fifths  of  an  inch.  Ten  centi- 
meters make  a  decimeter  (dm.)  (about  4  in.),  and  ten  deci- 
meters make  a  meter  (39.37  in.,  or  about  1.1  yd.).  We  may 
summarize  these  facts  in  the  following  reference  table : 

f  1  millimeter  =  0.03937  inches 
\linch  =  2.54  centimeters 

10  millimeters  =  1  centimeter  (0.3937  in.,  or  nearly  J  in.) 
10  centimeters  =  1  decimeter  (3.937  in.,  or  nearly  4  in.) 

10  decimeters  =  1  meter  (39.37  in.,  or  nearly  3-^  ft.) 


LINEAR  MEASUREMENT  29 

24.  Advantages  of  the  metric  system.  One  of  the  advan- 
tages of  the  system  is  that  the  value  of  the  fractional  part 
of  the  meter  is  more  apparent  than  a  corresponding  deci- 
mal part  of  a  yard.  Thus,  if  we  say  a  street  is  12.386  yd. 
wide,  the  decimal  0.386  tells  us  nothing  about  the  smaller 
divisions  of  a  yardstick  that  enter  into  this  number.  At 
best  we  would  probably  say  that  it  is  something  over  one 
third  of  a  yard.  On  the  other  hand,  if  we  say  that  a  road 
is  12.386  m.  wide,  we  know  at  once  that  the  road  is  12  m. 
3  dm.  8  cm.  6  mm.  wide.  This  last  statement  is  far  more 
definite  to  one  who  has  had  .a  little  practice  with  the 
metric  system. 

Obviously  the  advantages  of  the  metric  system  lie  in 
the  fact  that  ten  line  segments  of  any  unit  are  equal  to 
one  of  the  next  larger.  In  contrast  to  this  fact  the  multi- 
pliers .of  our  system,  though  they  may  seem  familiar,  are 
awkward.  Thus,  there  are  1 2  in.  in  a  foot,  3  ft.  in  a  yard, 
5^  yd.  in  a  rod,  1760yd.  in  a  mile,  etc. 

HISTORICAL  NOTE.  It  is  probable  that  most  of  the  standard  units 
of  length  were  derived  from  the  lengths  of  parts  of  the  human  body 
or  other  equally  familiar  objects  used  in  measuring.  Thus,  we  still 
say  that  a  horse  is  so  many  hands  high.  The  yard  is  supposed  to 
have  represented  the  length  of  the  arm  of  King  Henry  I.  Nearly  all 
nations  have  used  a  linear  unit  the  name  of  which  was  derived  from 
their  word  for  foot. 

During  the  French  Revolution  the  National  Assembly  appointed 
a  commission  to  devise  a  system  that  would  eliminate  the  inconven- 
ience of  existing  weights  and  measures.  The  present  metric  system 
is  the  work  of  this  commission. 

This  commission  attempted  to  make  the  standard  unit  one  ten- 
millionth  part  of  the  distance  from  the  equator  to  the  north  pole 
measured  on  the  meridian  of  Paris.  Since  later  measurements  have 
raised  some  doubt  as  to  the  exactness  of  the  commission's  determina- 
tion of  this  distance,  we  now  define  the  meter  not  as  a  fraction  of 
the  earth's  quadrant,  but  as  the  distance,  at  the  freezing  temperature, 


30  GENERAL  MATHEMATICS 

between  two  transverse  parallel  Jim-.*  ruled  on  a  bar  of  platinum-iridium 
which  is  kept  at  the  International  Butqau  of  Weight.*  and  Measures,  at 
Sevres,  near  Paris. 

25.  Application  of  the  metric  scale.    This  article  is  in- 
tended to  give  practice  in  the  use  of  the  metric  system. 

EXERCISES 

1.  With  a  ruler  whose  edge  is  graduated  into  centimeters 
measure  the  segments  AB  and  CD  in  Fig.  7. 


FIG.  7 

2.  Measure  the  length  and  width  of  your  desk  with  a  centi- 
meter ruler.   Check  the  results  with  those  of  Exs.  2  and  3, 
Art.  22. 

3.  Estimate  the  length  of  the  room   in  meters  and  then 
measure  the  room  with  a  meter  stick.    If  a  meter  stick  is  not 
available,  use  a  yardstick  and  translate  into  meters. 

4.  Turn  to  some  standard  text  in  physics   (for  example, 
Millikan  and  Gale,  pp.  2  and  3)  and  report  to  the  class  on  the 
metric  system. 

5.  Refer  to  an  encyclopedia  and  find  out  what  you  can 
about  the  "  standard  yard  "  kept  at  Washington. 

26.  Practical  difficulty  of  precise  measurement.  In  spite 
of  the  fact  that  measuring  line  segments  is  a  familiar 
process  and  seems  very  simple,  it  is  very  difficult  to  meas- 
ure a  line  with  a  high  degree  of  accuracy.  The  following 
sources  of  error  may  enter  into  the  result  if  we  use  a 
yardstick  :  (1)  the  yardstick  may  not  be  exactly  straight  ; 
(2)  it  may  be  a  little  too  long  or  too  short  ;  (3)  it  may 
slip  a  little  so  that  the  second  position  does  not  begin  at 


LINEAR  MEASUREMENT 


31 


the  exact  place  where  the  first  ended ;  (4)  the  edge  of 
the. yardstick  may  not  always  be  along  the  line  segment; 
(5)  the  graduated  scale  used  for  reading  feet,  inches,  and 
fractional  parts  of  inches  may  not  be  correct.  Nor  do  we 
eliminate  these  errors  by  using  other  measuring  devices. 
For  example,  a  tape  line  tends  to  stretch,  but  contracts  if 
wet,  while  a  steel  tape  is  affected  by  heat  and  cold. 

From  the  preceding  discussion  it  is  apparent  that  a 
measurement  is  always  an  approximation.  The  error  can 
be  decreased  but  never  wholly  eliminated. 

EXERCISES 

1.  Suppose  you  have  measured  a  distance  (say  the  edge  of  your 
desk)  with  great  care  and  have  found  it  to  be  equal  to  2  ft.  7f  in. 
Is  this  the  exact  length  of  the  desk  ?  Justify  your  answer. 

2.  If  you  were  to  repeat  the  measurement  with  still  greater 
care,  making  use  of  a  finer-graduated  scale,  is  it  likely  that 
you  would  find  exactly  the  same  result 

as  before  ? 

3.  If  you  were   asked   to  measure 
the   length   of   your    classroom,  what 
would  you  use  ?   Why  ? 


27.  The  compasses.  A  pair  of  com- 
passes (Fig.  8)  is  an  instrument 
that  may  be  used  in  measuring  line 
segments.  Since  the  use  of  com- 
passes greatly  decreases  some  of  the 
common  errors  in  measuring  that 
have  been  pointed  out,  and  is  con- 
sequently very  useful  in  many  forms  of  drawing  which 
require  a  high  degree  of  accuracy,  it  will  be  helpful  if  the 
student  learns  to  use  the  compasses  freely. 


FIG.  8.    A  PAIR  OF 
COMPASSES 


32  GENERAL  MATHEMATICS 

28.  Measuring  a  line  segment  with  the  compasses.  To 
measure  the  line  segment  AB  in  Fig.  9  with  the  compasses, 
place  the  sharp  points  of  the  compasses  on  A  and  B.  Turn 


FIG.  9 

the  screw  which  clamps  the  legs  of  the  compasses.  Then 
place  the  points  on  the  marks  of  the  ruler  and  count  the 
number  of  inches  or  centimeters  between  them. 

EXERCISES 

1.  With  the  compasses  measure  AB  in  Fig.  9,  in  inches. 

2.  With  the  compasses  measure  AB  in  Fig  9,  in  centimeters. 

3.  Multiply  the  result  of  Ex.  1  by  2.54.  What  do  you  observe  ? 

4.  Estimate  the  number  of  centimeters  in  the  length  of  this 
page.    Measure  the  length  of  this  page  with  the  compasses  and 
compare  with  your  estimate. 

29.  Squared  paper.    Squared  paper  is  another  important 
device  which  is  often  useful  in  measuring  line  segments. 

Squared  paper  is  ruled  either    A  B 

to  inches  and  fractions  of  an 
inch  (used  by  the  engineer)  or 
to  the  units  of  the  metric  scale. 
A  sample  part  of  a  sheet  is 
shown  in  Fig.  10.  The  method 
of  measuring  with  squared  pa- 
per is  practically  the  same  as 
measuring  with  the  compasses 
and  ruler.  Thus,  to  measure 
the  line  segment  AB  in  Fig.  10 
place  the  sharp  points  of  the  compasses  on  A  and  B. 
Clamp  the  compasses.  Place  the  sharp  points  on  one  of 


FIG.  10.    HOAV  A  LINE  SEGMENT 

MAY  BE  MEASURED  BY  THE  USE 

OF  SQUARED  PAPER 


LINEAR  MEASUREMENT 


33 


the  heavy  lines,  as  at  E  and  D.  Each  side  of  a  large 
square  being  1  cm.,  count  the  number  of  centimeters  in 
ED  and  estimate  the  remainder  to  tenths  of  a  centimeter. 
Thus,  in  Fig.  10  the  segment  ED  equals  2.9  cm. 

EXERCISES 

1.  Draw  a  line  segment  and  measure  its  length  in  centi- 
meters by  the  use  of  squared  paper. 

2.  Why  do  you  find  it  convenient  to  place  one  of  the  sharp 
points  of  the  compasses  tyhere  two  heavy  lines  intersect  ? 

3.  What  are  the   advantages  of   measuring  with  squared 
paper  over  measuring  with  a  ruler  ? 

30.  Measuring  a  length  approximately  to  two  decimal 
places.  By  increasing  the  size  of  the  unit  of  measurement 
we  may  express  the  result  with  approximate  accuracy  to 
two  decimal  places.  In  Fig.  11  let  MN  (equal  to  ten  small 
units,  or  2  cm.,  of  A  •  % 

the  squared  paper)     ' 
be   the   unit.     As 
before,  lay  off  AB 
upon  the  squared 
paper     with      the 
compasses  (CD  in 
the  new  position). 
Now  EF  (a  small 
unit)  equals  0.1  of 
a  unit,  and  0.1  of  EF  equals  0.01  of  the  unit  MN.  Reading 
the  units  in  the  line  CD,  we  are  sure  the  result  is  greater 
than  2.7,  for  the  crossing  line  is  beyond  that.   But  it  is  not 
2.8.    Why?    Now  imagine  a  small  unit,   as  EF,  divided 
into  tenths  and  estimate  the  number  of  tenths  from  the 
end  of  the  twenty-seventh  small  unit  to  D.    This  appears 


N 


Unit 


FIG.  11 


GENERAL  MATHEMATICS 


to  be  0.4,  but  this  is  0.04  of  a  unit;,  hence  CD  equals 
2.74  units.  This  means  that  CD  is  2.74  times  as  long  as 
the  line  MN.  Of  course  the  4  is  only  an  approximation, 
but  it  is  reasonably  close. 


\ 


Kt- 


EXERCISES 

1.  In  Fig.  12  measure  to  two  decimal  places  the  segments 
CD,  DE,  and  EC.    Compare  the  results  of  your  work  with 

that  of  the  other  members  of 

the  class. 

2.  Is  the  result  obtained  by 
the   method    of  Art.  30   more 
accurate    than    the    result    ob- 
tained by  using  1  cm.  as  a  unit 
and  claiming  accuracy  to  only 

one  decimal  place  ? 

FIG.  12 

31.  Equal    line    segments. 

When  the  end  points  of  one  segment,  as  a  in  Fig.  13,  coin- 
cide with  (exactly  fit  upon)  the  ends  of  another  segment, 
as  6,  the  segments  a  and  b  are  said  to  be 
equal.  This  fact  may  be  expressed  by 
the  equation  a  =  b. 

32.  Unequal  segments;  inequality.    If 
the  end  points  of  two  segments,  as  a  and  J, 

cannot  possibly  be  made  to  coincide,  the  segments  are  said  to 
be  unequal.   This  is  written  a  =£  b  (read  "  a  is  not  equal  to  6"). 

The  statement  a  3=  bis  called  . a | 

an  inequality.  In  Fig.  14  seg- 
ment a  is  less  than  segment  b 
(written  a  <  i),  and  seg- 
ment c  is  greater  than  seg- 
ment b  (written  c  >  6). 


I 1 

I b- 1 

FIG.  13.  EQUAL  LINE 
SEGMENTS 


V 


FIG.  14.   UNEQUAL  LINE  SEGMENTS 


LINEAR  MEASUREMENT  35 

In  the  preceding  equation  and  inequalities  we  need  to 
remember  that  the  letters  a,  b;  and  c  stand  for  the  length 
of  the  segments.  They  represent  numbers  which  can  be 
determined  by  measuring  the  segments. 

33.  Ratio.  This  article  will  show  that  ratio  is  a  funda- 
mental notion  in  measurement. 

INTRODUCTORY  EXERCISES 
(Exs.  1-4  refer  to  Fig.  15) 

1.  Measure  the  segment  a  accurately  to  two  decimal  places. 

2.  Measure  the  segment  b 

accurately    to    two    decimal    | 1 

places. 

3.  What  part  of  b  is  a  ?  \—  —I 

4.  Find  the  quotient  of  b  FlG  15 
divided  by  a. 

The  quotient  of  two  numbers  of  the  same  kind  is  called 
their  ratio.  The  ratio  is  commonly  expressed  as  a  fraction. 
Before  forming  the  fraction  the  two  quantities  must  be 
expressed  in  terms  of  the  same  unit;  for  example,  the 
ratio  of  2  ft.  to  5  in.  is  the  ratio  of  24  in.  to  5  in.,  that 
is,  -2^4-.  The  unit  of  measure  is  1  in.  Obviously  there  is 
no  ratio  between  quantities  of  different  kinds ;  for  exam- 
ple, there  is  no  ratio  between  7  gal.  and  5  cm. 

It  should  now  be  clear  that  every  measurement  is  the 
determination  of  a  ratio  either  exact  or  approximate.  Thus, 
when  we  measure  the  length  of  the  classroom  and  say  it>  is 
10  m.  long,  We  mean  that  it  is  ten  times  as  long  as  the 
standard  unit,  the  meter ;  that  is,  the  ratio  is.  -L<>.. 

•    •  ••  -  -     -   -  •' •  -.  .-  -       -: 


36 


GENERAL  MATHEMATICS 


EXERCISES 

1.  The  death  rate  in  Chicago  in  a  recent  year  was  16  to  1000 
population.    Express  this  ratio  as  a  fraction. 

2.  An  alloy  consists  of  copper  and  tin  in  the  ratio  of  2  to  3. 
What  part  of  the  alloy  is  copper  ?    What  part  is  tin  ? 

3.  A  solution  consists  of  alcohol  and  water  in  the  ratio  of 
3  to  6.    What  part  of  the  solution  is  water  ? 

4.  The  ratio  of  weights  of  equal  volumes  of  water  and  cop- 
per is  given  by  the  fraction  —  •    How  many  times  heavier  is 
copper  than  water  ? 

5.  Water  consists  of  hydrogen  and  oxygen  in  the  ratio  of 
1  to  7.84.    Express  this  ratio  as  a  decimal  fraction. 

34.  Sum   of  two    segments ;  geometric    addition.    It   is 

possible  to  add  two  line  segments  by  the  use  of  compasses. 
Thus,  in  Fig.  16  if  the  segment  a  is  laid  off  on  the  number 
scale  of  squared  paper  from  point  A  to  point  B  and  if  in 
turn  b  is  laid  off  on  the  same  line  from  B  to  C,  then  the 


FIG.  16.    GEOMETRIC  ADDITION  OF  LINE  SEGMENTS 

sum  of  these  lines  can  be  read  off  at  once.  In  Fig.  16  the 
sum  is  5.4  cm.  The  segment  AC  is  the  sum  of  a  and  b. 
Very  often  in  construction  work  we  are  not  concerned 
about  either  the  length  of  the  segments  or  their  sum.  In 
that  case  lay  off  the  segments  as  above  on  a  working  line 
and  indicate  the  sura  of  a  and  b  as  a  •+-  b.  Addition  per- 
formed by  means  of  the  compasses  is  a  geometric  addition. 


LINEAR  MEASUREMENT 


37 


FIG.  17 


EXERCISES 

1.  In  Fig.  17  find  the  sum  of  a,  b,  and  c  on  a  working 

line.    Indicate  the  sum. 

. « . 

2.  In  Fig.  17  add  the  seg- 

ments  a,  b,  and  c  on  the  scale    i - 1 

line  of  a  sheet  of  squared  c 

paper.  Express  the  value  of 
a  +  b  +  c  in  centimeters. 

3.  On  a  working  line  draw 
one  line  to  indicate  the  num- 
ber   of    yards    of    fencing 
needed  for  the,  lot  in  Fig.  18. 

4.  In  Fig.  19  the  whole 
segment   is    denoted   by   c. 
Show     by     measuring     on 
squared  paper  that  c  =  a  +  b. 

5.  In  Fig.  19  what  rela- 
tion exists   between  c  and 
either  a  or  b?  Why? 


FIG.  18 


FIG.  19 


35.  Axioms.    Exs.  4  and  5,  above,  illustrate  the  following 
two  axioms : 

VI.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

VII.  The  whole  is  greater  than  any  one  of  its  parts. 


EXERCISES 

1.  Draw  the  segments  a  =  2.3  cm.;  b  =  3.2  cm.;  c  =  1.3cm. 
Draw  the  sum  a  +  b  +  c. 

2.  Let  a,  b,  and  c  denote  three  line  segments.    Draw  a  line 
segment  to  represent  2a  +  3b  +  c ;  to  represent  4 a  -f  b -f  2 c ; 
to  represent  a  -\-  3  b  -j-  4  c. 


38  GENERAL  MATHEMATICS 

3.  In  Fig.  20,  if  a,  b,  and  c  are  three  consecutive  segments  on  a 
straight  line,  such  that  a  =  c,  show    | , I | 

by  measuring  that  a  +  b  =  I  +  c. 

FIG.  20 

4.  Show  with  out  measuring  that 

a  +  b  =  b  +  c.  What  axiom  does  this  fact  illustrate  ?  Quote 
the  axiom. 

36.  Order  of  terms  in  addition.    The  fact  that  We  get  the 
same  sum  when  we  lay  off  a  segment  a  and  then  add  b 
that  we  do  when  we  lay  off  b  first  and  then  add  a  is  a 
geometric  illustration  of  the  truth  of  the  commutative  law. 
This  law  asserts  that  the  value  of  a  sum  does  not  change 
when  the  order  of  the  addends  is  changed.    In  the  first  chap- 
ter we  illustrated  this  principle  by  a  familiar  experience 
from  arithmetic,  as  2  +  5-1-4  =  2  +  4-}- 5. 

EXERCISES 

1.  Illustrate  the  validity  of  the  commutative  law  by  a  fact 
from  your  everyday  experience. 

2.  Add  in  the  most  advantageous  way,  using  the  commutative 
law :  .376  +  412  +  124 ;  2187  +  469  +  213 ;  36  +  142  + 164. 

37.  Difference  of  two  line  segments.    The  difference  of 
two  line  segments  may  also  be  found  with  the  compasses. 
To  find  out  how  much  greater      | . 1 

the  segment  c  is  than  the  seg- 
ment b  we  lay  off  the  segment  c 
on  a  working  line  (Fig.  21)  from 
A  to  C,  then  lay  the  segment  A.  D  b  c 

b  backward  from  C  toward  A.      FIG.  21.    GEOMETRIC  SUBTRAC- 

Then  the   difference    between 

the  segments  b  and  c  is  expressed  by  the  segment  AD.  In 
equation  form  this  may  be  written  AD  =  c  —  b.  Illustrate 
this  method  by  comparing  the  lengths  of  two  pencils. 


c-b 


LINEAR  MEASUREMENT  39 

EXERCISES 

1.  Transfer  the  line  segments  of  Fig.  21  to  squared  paper 
and  express  the  difference  between  c  and  b  in  centimeters. 

2.  Subtract  a  line  segment  3.5  cm.  long  from  one  6  cm.  long. 

3.  In  Fig.  22  the  line  segment  AB  equals  the  line  segment 
MX.  Show  by  measuring  that  M  h         N 

AB  -  MB  =  MN  -  MB.  \ h^-+ 1 

A  B 

4.  Ex.  3  is  simpler  if  we  ^IG  22 
write  the   fact   in   algebraic 

form,  using  the  small  letters.    Thus,  a  +  c  =  b  +  c.    How  would 
you  show  that  a  =  b  ? 

5.  What  axiom  of  the  first  chapter  is  illustrated  by  Exs.  3 
and  4  ?   Quote  the  axiom. 

6.  If  a  =  3  cm.,  b  =  2  cm.,  and  c  —  1  cm.,  construct  a  line 
segment  representing  2 a  +  3 b  —  c ;  representing  5a  —  2b  +  c. 

7.  If  a,  b,  and  c  represent  the  length  of  three  respective  seg- 
ments, construct  3  <z  -f-  2  #  —  c ;  4  a  +  2  £  —  2  c ;  5 a  —  2 6.+  3 c. 

8.  How  long  would  each  of  the  segments  constructed  in 
Ex.  7  be  if  a  =  5,  b  =  4,  and  c  =  3  ? 

38.  Coefficient.  The  arithmetical  factor  in  the  term  2# 
is  called  the  coefficient  of  the  literal  factor  #.  When  no 
coefficient  is  written,  as  in  #,  we  understand  the  coeffi- 
cient to  be  1.  Thus,  x  means  1  x.  The  coefficient  of  a 
literal  number  indicates  how  many  times  x 

the  literal  number  is  to  be  used  as  an  ~ ~ 

addend;  thus,  5 a:  means  x+x+x+z+x 
and  can  be  expressed  by  the  equation  5x  =  x+  x+x-\-x+x. 
Written  in  this  form  we  see  that  the  use  of  a  coefficient 
is  a  convenient  method  of  abbreviating.  Geometrically  a 
coefficient  may  be  interpreted  as  follows:  Let  x  be  the 
length  of  the  segment  in  Fig.  23.  Then  the  5  in  5x  indi- 
cates that  the  line  segment  x  is  to  be  laid  off  five  times 


40  GENERAL  MATHEMATICS 

consecutively  on  a  working  line.  5.r  expresses  the  sum 
obtained  by  this  geometric  addition.  Find  this  sum.  Usu- 
ally the  term  "coefficient"  means  just  the  arithmetical  factor 
in  a  term,  though  in  a  more  general  sense  the  coefficient 
of  any  factor,  or  number  in  a  term,  is  the  product  of  all  the 
other  factors  in  that  term.  Thus,  in  3  aby  the  coefficient 
of  y  is  3  ab,  of  by  is  3  a,  of  aby  is  3. 

EXERCISE 

Give  the  coefficient  in  each  of  the  following  terms  :  3  b  ; 
x     7x     8#     9  a-. 


39.  Polynomials.    An  algebraic  number  consisting  of  two 
or  more  terms  (each  called  a  monomial),  as  bx  +  %  y  —  2  2, 
is  a  polynomial.    The  word  "polynomial"  is  derived  from 
a  phrase  which  means  many  termed.    A  polynomial  of  two 
terms,  as  5  x  +  3  #,  is  a  binomial.    A  polynomial  of  three 
terms,  as  2a-f3J4-4c,  is  a  trinomial. 

EXERCISE 

Classify   the   following   expressions   on   the   basis    of   the 
number  of  terms  : 

(a)  2  m  +  3  n  —  5  x  +  r.  (c)  6  x  +  2  y. 

(b)  6x.  (d)  a  +  2&  +  3. 

40.  Algebraic  addition  of  similar  terms.    In  simple  prob- 
lems we  have  frequently  added  similar  terms.    We  shall 
now  review  the  process  by  means  of  the  following  example 

•in  order  to  see  clearly  the  law  to  be  used  in  the  more 
complicated  additions  : 

Add  4:r-!-3a:  +  2:r. 

Solution.   4  x  can  be  considered  as  the  sum  of  four  segments  each 
x  units  long. 

Therefore  4x  =  z  +  z  +  r  +  x. 


LINEAR  MEASUREMENT  41 

Similarly,  3  x  =  x  +  x  +  x, 

and  2x  =  z  +  x. 

Adding, 

4:X  +  3x  +  2x=:x  +  x  +  x  +  x  +  x  +  x  +  x  +  x  +  x,  or  9  z. 
Hence       .  4z  +  3z  +  2z  =  9z. 

The  preceding  example  illustrates  the  law  that  the  sum 
of  two  or  more  similar  monomials  is  a  monomial  whose  coeffi- 
cient is  the  sum  of  the  coefficients  of  the  given  monomials  and 
which  has  the  same  literal  factor  as  the  given  monomials. 

The  advantages  of  adding  numbers  according  to  this 
law  may  be  seen  by  comparing  the  two  solutions  of  the 
following  problem : 

The  tickets  for  a  school  entertainment  are  sold  at  250  each 
by  a  cominittee  of  five  students,  A,  B,  C,  D,  and  E.  A  reports 
38  sold;  B,  42;  C,  26;  D,  39;  and  E,  57.  At  the  door  173 
tickets  are  sold.  Find  the  total  receipts. 

Solution  I. 


A's  receipts, 

38 

x 

$0.25  = 

$9.50 

B's  receipts, 

42 

x 

0.25  = 

10.50 

C's  receipts, 

26 

X 

0.25  = 

6.50 

D's  receipts, 

39 

X 

0.25  = 

9.75 

E's  receipts, 

57 

X 

0.25  = 

14.25 

Door  receipts, 

173 

X 

0.25  = 

43.25 

Total  receipts, 

$93.75 

Since  the  common  factor  in  the  above  multiplication  is 
25,  a  simpler  solution  is  obtained  if  the  numbers  of  tickets 
sold  (coefficients)  are  first  added  and  placed  before  the 
common  factor ;  thus, 

Solution  II.     A's  receipts,  38  x  $0.25 


B's  receipts, 

42  x 

0.25 

C's  receipts, 

26  x 

0.25 

D's  receipts, 

39  x 

0.25 

E's  receipts, 

57  x 

0.25 

Door  receipts, 

173  x 

0.25 

Total  receipts, 

375  x 

0.25 

=  $03.75 

42  GENERAL  MATHEMATICS 

EXERCISES 

1.  Tickets  were  sold  at  c  cents  ;  A  sells  12;  B,  15;  C,  36; 
and  D,  14.    There  were  112  tickets  sold  at  the  gate.   Find  the 
total  receipts. 

2.  Express  as  one  term  3-7  +  5-7-f4-7. 

NOTE.  A  dot  placed  between  two  numbers  and  halfway  up  in- 
dicates multiplication  and  is  read  "times."  Do  not  confuse  with 
the  decimal  point. 

3.  Can  you  add  3  •  7  +  14  •  2  -f  5  •  4  by  the  short  cut  above? 

4.  Indicate  which  of  the  following  sums  can  be  written  in 
the  form  of  monomials  :  3x  +  5a;;  ±x  +  7  x  +  3;  13  +  5  -|-  3  ; 
3  «  +  2  1  +  4. 

5.  Add  as  indicated  :  (a)  3  x  +  20  x  +  17  x  +  9  x  +  ~x  +  3  x  ; 
(b)  3y  +  y  +  15y+ily  +  2y;  (c)  9*  +  3s  +  3s  +  4  .s  -j-  2s  ; 


6.  The  school's  running  track  is  /  feet.  While  training,  a 
boy  runs  around  it  five  times  on  Monday,  six  on  Tuesday,  ten 
on  Wednesday,  seven  011  Thursday,  six  on  Friday,  and  nine  on 
Saturday  morning.  How  many  feet  does  he  run  during  the  week  ? 

41.  Subtraction  of  similar  monomials.    The  law  in  sub- 
traction is  similar  to  the    law  in   addition   and  may  be 
illustrated  as  follows: 
Subtract  2  x  from  5  x. 
Solution.  5x  =  x  +  x  +  x  +  x  +  x. 

2  x  =  x  +  x. 
Subtracting  equal  numbers  from  equal  numbers, 

5  x  —  2  x  =  x  +  x  +  x,  or  3  x. 
Hence  5x  —  2x  =  3  x. 

The  preceding  example  illustrates  the  law  that  tJie  differ- 
ence of  two  similar  monomials  is  a  monomial  having  a  coefficient 
equal  to  the  difference  of  the  coefficients  of  the  given  monomials 
and  having  the  same  literal  factor. 


LINEAR  MEASUREMENT 


43 


EXERCISES 

1.  Subtract  3  b  from  146. 

2.  Write  the  differences  of  the  following  pairs  of  numbers  as 
monomials:  lOce  —  3ic;  13x  —  5x;  12z  —  3z;  VI k  —  5k;  2.68  r 
-0.27/-;  1.03a-0.08a;  fe-Je;  f*  — i*. 

3.  The  following  exercises  require  both  addition  and  sub- 
traction.   Write  each  result  as  a  single  term  :  <ix-}-6x  —  2x; 
13x  —  2x  +  3x;  11.5c  + 2.3c  -  c;  Ja  +  f  a  —  J  a. 

42.  Triangle;  perimeter.  If  three  points,  as  A,  B,  and 
(7  (Fig.  24),  are  connected  by  line  segments,  the  figure 
formed*  is  a  triangle.  The 
three  points  are  called  ver- 
tices (corners)  of  the  triangle, 
and  the  three  sides  a,  6,  and  A 
c  are  the  sides  of  the  -tri- 
angle. The  sum  of  the  three  sides,  as  a  -f-  b  +  c  (the 
distance  around),  is  the  perimeter  of  the  triangle. 

EXERCISES 

1.  A  yard  has   the   form   of   an  equal-sided   (equilateral) 
triangle,  each  side  being  x  rods  long.    How  many  rods  of 
fence  will  be  needed  to  inclose  it  ? 

2.  What  is  the  sum  of  the  sides 
of  a  triangle  (Fig.  25)  whose  sides 
are  2  x  feet,  2  x  feet,  and  3  x  feet 
long?    Express  the  sum   as  a  cer- 
tain number  of  "times  x. 

3.  What  is  the  sum  of  the  three  sides  3b,  4b,  and  6b  of  a 
triangle  ?  Express  the  result  as  one  term. 

4.  What  is  the  perimeter  of  a  triangle  whose  sides  are  2x, 
8  x}  and  9  x  ?   Let  p  be  the  perimeter ;  then  write  your  answer 
to  the  preceding  question  in  the  form  of  an  equation. 


2x 


3x 
FIG.  25 


44 


GENERAL   MATHEMATICS 


43.  Polygons.  A  figure,  as  ABODE  (see  Fig.  26),  formed 
by  connecting  points,  as  A,  J5,  C,  D,  and  E,  by  line 
segments,  is  a  polygon. 
The  Greek  phrase  from 
which  the  word  "polygon" 
is  derived  means  many  cor- 
nered.  Polygons  having  3, 
4,  5,  6,  8,  10,  •  •  .,  n  sides 
are  called  triangle,  quadri- 
lateral, pentagon,  hexagon, 
octagon,  decagon,  -  -  .,  n-gon  FlG  26.  A  POLYGON 

respectively.    The  sum  of 

the  sides  of  a  polygon  is  its  perimeter.    When  all  the 
sides  of  a  polygon  are  equal  it  is  said  to  be  equilateral. 


EXERCISES. 


1.  What  is  the  perimeter  of  each  of  the  polygons  in  Fig.  27? 
In  each  case  express  the  result  in  the  form  of  an  equation ; 
thus,  for  the  first  quadrilateral  p  =  12  x. 


IX 


FIG.  27 


2.  Show  by  equations  the  perimeter  p  of  the  polygons  in 
Fig- 28. 

3.  Show  by  equations  the 
perimeter  of  an  equilateral 
quadrilateral  whose  side  is 
11;    9;    s;    x;    b;    z\    2e; 
9  +  3;  a  +  5;  a  +  d;  x  +  7 : 
x-fy. 


10        ,2 

c 

d 

FIG.  28 

LINEAR  MEASUREMENT  45 

4.  Name  the  different  figures  whose  perimeters  might  be 
expressed  by  the  following  equations : 

p  =  3  s,     p  =  5  s,     j)  =  7  s,     p  =  9  s,       p  =  12  s,     p  =  20  s, 
p  =  4  s,     p  =  6  s,     p  =  8  s,     p  =  10  s,     p  =  15  s,     p  =  ns. 

5.  Assume  that  all  the  figures  in  Ex.  4  are  equilateral. 
Find  out  how  many  of  your  classmates  can  give  the  name  of 
each  polygon. 

6.  Assume  that  at  least  six  of  the  polygons  in  Ex.  4  are 
not  equilateral.    Sketch  the  figures  of  which  the  given  equa- 
tions may  be  the  perimeters. 

7.  What  is  the  perimeter  of  each  figure  of  Ex.  4  if  s  =  2  in.  ? 
ifs  =  3cm.?  if  s  =  4yd.?  ifs  =  5ft.? 

8.  Determine  the  value  of  s  in  the  equations  p  =  3  s,  p  =  4  s, 
p  =  5s,  p  =  6s,  p  =  10s,  and  p  =  15s   if  in  each  case  the 
perimeter  is  120  in. 

9.  What  is  the  side  of  an  equilateral  hexagon  made  with  a 
string  144  in.  long  ?    (Use  all  the  string.) 

10.  Show  by  sketches  polygons  whose  perimeters  are  ex- 
pressed by  p  =  8  a  +  6  ;    by^?  =  4«  +  12;    by^?  =  66  +  6a; 
by^>  —  4a  +  25;  by  p  =  3a  -\-  2b. 

11.  Find  the  value  of  the  perimeters  in  Ex.10  if  a  =  3 
and  b  =  2 ;  if  a  =  5  and  b  =  3  ;  if  a  =  1  and  b  =  5. 

12.  If  x  =  2  and  y  —  3,  find  the  value  of  the  following 
expressions :  3x  +  ?/ ;  3x  —  y;  3x  —  2 ?/ ;  2x  —  3|- ;  4 a;  —  2-^  ?/ ; 
2.25 x-y;  2.27 x-  1.12 y. 

SUMMARY 

44.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  line  segment,  point,  measurement 
of  length,  unit  segment,  standard  unit,  ratio,  metric  system, 
coincide,  intersect,  equal  segments,  unequal  segments,  com- 
mutative law,  coefficient,  polynomial,  binomial,  trinomial, 


46  GEKEKAL  MATHEMATICS 

triangle,  vertex  of  a  polygon,  vertices  of  a  polygon,  perim- 
eter, sides  of  a  triangle,  polygon,  quadrilateral,  pentagon, 
hexagon,  octagon,  decagon,  n-gon,  equilateral. 

45.  Axioms.    The  following  axioms  were  illustrated: 

VI.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

VII.  The  whole  is  greater  than  any  of  its  parts. 

46.  The  following  instruments  have  been  used  in  measur- 
ing line  segments :  the  ruler,  the  compasses,  and  squared  paper. 

47.  The  following  symbols  were  used :.  =£  meaning  does 
not  equal;    <  meaning  is  less  than;    >  meaning  is  greater 
than ;  and  a  dot,  as  in  3  •  5,  meaning  times,  or  multiplied  by. 

48.  A  point  is  determined  by  two  intersecting  lines. 

49.  The  metric  system  has  certain  advantages  over  our 
English  system. 

50.  The  practical  difficulty   of  precise   measuring   has 
been  pointed  out.    Five  possible  errors  were  enumerated. 
Measurement  implies  the  determination  of  a  ratio. 

51.  The  sum  of  two  segments  was  found  with  the  com- 
passes.   A  law  was  discovered  to  serve  as  a  short  cut  in 
algebraic  addition. 

52.  The  difference  of  two  segments  was  found  with  the 
compasses  and  the  law  for  algebraic  subtraction  was  stated. 

53.  The  Addition  and  Subtraction  laws  of  Chapter  I 
were  illustrated  geometrically. 

54.  The  perimeter  of  a  figure  may  be  expressed  by  an 
equation. 

55.  The  chapter  has  taught  how  to  find  the  value  of  an 
algebraic  number  when  the  value  of  the  unknowns  are 
given  for  a  particular  cage;  for  example,  how  to  find  the 
value  of  3 x  +  2 y  when  x-  ==  1  and  #==2. 


CHAPTER  III 

PROPERTIES  OF  ANGLES 

56.  Angle.  If  a  straight  line,  as  OX  in  either  of  the 
drawings  of  Fig.  29,  rotates  in  a  plane  about  a  fixed  point, 
as  0,  in  the  direction  indicated  by  the  arrowheads  (counter- 
clockwise) until  it  reaches  the  position  OT,  it  is  said  to 


x 


FIG.  29.    ILLUSTRATING  THE  DEFINITION  OF  AN  ANGLE 


turn  through  the  angle  XOT.  Thus,  an  angle  is  the  amount 
of  turning  made  by  a  line  rotating  about  a  fixed  point  in  a 
plane  (flat  surface}.  Note  that  as  the  rotation  continues, 
the  angle  increases. 

57.  Vertex.    The  fixed  point  0  (Fig.  29)  is  called  the 
vertex  of  the  angle.    (The  plural  of  "vertex"  is  "vertices.") 

47 


48  GENERAL  MATHEMATICS 

58.  Initial  side ;  terminal  side.    The  line  OX  (Fig.  29) 
is  called  the  initial  side  of  the  angle.    The  line  OT  is  called 
the  terminal  side  of  the  angle. 

59.  Symbols  for  "  angle."   The  symbol  for  "angle"  is  Z; 
for  "  angles,"  A     Thus,  "  angle  XOT"  is  written  /^XOT. 

60.  Size  of  angles.   From  the  definition  of  an  angle  given 
in  Art.  56  we  see  that  it  is  possible  for  the  line  OX  to  stop 
rotating  (Fig.  29)  so  that  the  angle  may  contain  any  given 
amount  of  rotation  (turning). 

EXERCISE 

Draw  freehand  an  angle  made  by  a  line  OX  which  has 
rotated  one  fourth  of  a  complete  turn ;  one  half  of  a  complete 
turn ;  three  fourths  of  a  complete  turn ;  one  complete  turn ; 
one  and  one-fourth  complete  turns. 

61.  Right  angle ;    straight  angle  ;    perigon.     If   a  line 
rotates    about    a  fixed    point  in   a  plane  so   as  to  make 

T       one  fourth   of  a  complete  turn,  the    angle   formed 
is  called  -a  right  angle  (rt.  Z)  (see  Fig.  30,  (a)). 


-X 


O 
(a)  Right  Angle  (b)  Straight  Angle  (c)  Perigon 

FIG.  30.   THREE  SPECIAL  ANGLES 

If  the  line  makes  one  half  of  a  complete  turn,  the  angle 
formed  is  called  a  straight  angle  (st.  Z)  (see  Fig.  30,  (b))  ;  if 
the  line  makes  a  complete  turn,  the  angle  formed  is  called 
&  perigon  (see  Fig.  30,  (c)). 

EXERCISES 

1.  Draw  freehand  an  angle  equal  to  1  right  angle;  2  right 
angles  ;  3  right  angles  ;  4  right  angles. 

2.  Draw    freehand   an    angle    equal    to   1   straight    angle; 
2  straight  angles ;   1|-  straight  angles ;    2^  straight  angles. 


PKOPEKTIES  OF  ANGLES 


49 


3.  How  many  right  angles  are  there  in  a  half  turn  of  a 
rotating  line  ?   in  a  whole  turn  ?  in  5  turns  ?  in  3^  turns  ?  ' 

in  x  turns  ?    in  -  turns  ?    in  -  turns  ? 
2  4 

4.  How  many  straight  angles  are  there  in  a  half  turn  of  a 
rotating  line  ?  in  a  whole  turn  ?  in  1^  turns  ?  in  1|  turns  ? 

5x 

in  x  turns  ?  in  2x  turns  ?  in  -—  turns  ? 

4 

5.  Through  how  many  right  angles  does  the  minute  hand  of 
a  watch  turn  in  3  hr.  ?    in  3^  hr.  ?    in  4  hr.  ?    in  1\  hr.  ?    in 
x  hours  ?    in  15  min.  ?    in  30  min.  ?    in  5  min.  ?    in  10  min.  ? 

6.  How  many  straight  angles  are  there  in  a  perigon  ?  in 
a  right  angle  ?    in  10  right  angles  ?   in  y  right  angles  ? 

62.  Acute   angle ;   obtuse   angle ;   reflex   angle.    Angles 
are  further  classified  upon  the  basis  of  their  relation  to 

•  T 


FIG.  31.   ACUTE  ANGLE 


FIG.  32.    OBTUSE  ANGLE 


the  right  angle,'  the  straight  •  angle,  and  the  perigon.    An 

angle   less  than   a  right   angle  is   called   an  acute   angle, 

Fig.  31.  An  angle 

which   is   greater 

than  a  right  angle 

and   is   less   than-  'H 

a    straight    angle 

is  called  an  obtuse 

angle,  Fig.  32.  An 

angle  greater  than 

a    straight    angle 

and  less  than  a  perigon  is  called  a  reflex  angle,  Fig.  33. 


FIG.  33.    REFLEX  ANGLE 


GENERAL  MATHEMATICS 


EXERCISES 

4 

1.  Draw  an  acute  angle  ;  an  obtuse  angle  ;  a  reflex  angle. 

2.  Point  out,  in  the  classroom,  examples  of  right  angles ; 
of  obtuse  angles. 

C\ 
C 

D 


(a)  (b)  (c) 

FIG.  34.    ILLUSTRATING  THE  VARIOUS  KINDS  OF  ANGLES 

3.  In  the  drawings  of  Fig.  34  determine  the  number  of  acute 
angles  ;  of  right  angles  ;  of  obtuse  angles  ;  of  reflex  angles. 

63.  Notation  for  reading  angles.  There  are  three  common 
methods  by  which  one  may  denote  angles:  (1)  Designate 
the  angle  formed  by  two  lines  OX 
and  OT  as  the  "  angle  XOT"  or  the 
"  angle  TOX"  (Fig.  35).  Here  the 
first  and  last  letters  denote  points 
on  the  lines  forming  the  angle,  and 
the  middle  letter  denotes  the  point 
of  intersection  (the  vertex).  In  read- 
ing "  angle  XOT"  we  regard  OX  as 
the  initial  side  and  OT  as  the  terminal  side.  (2)  Denote  the 
angle  by  a  small  letter  placed  as  #  in  Fig.  36.  In  writing 
equations  this  method  is  the  B 

most  convenient.  (3)  Denote 
the  angle  by  the  letter  which 
is  written  at  the  point  of  in- 
tersection of  the  two  sides  of 
the  angle,  as  "  angle  A"  (Fig.  36).  This  last  method  is  used 
only  when  there  is  no  doubt  as  to  what  angle  is  meant. 


FIG.  35 


FIG.  36 


PKOPERTIES  OF  ANGLES 

EXERCISE 


51 


In  the  drawings  of  Fig.  37,  below,  select  three  angles  and 
illustrate  the  three  methods  of  notation  described  above. 


(b) 


FIG.  37 


64.  Circle.  If  a  line  OX  be  taken  as  the  initial  side 
of  an  angle  (see  Fig.  38)  and  the  line  be  rotated  one 
complete  turn  (a  peri- 
gon),  any  point,  as  P, 
on  the  line  OX  will 
trace  a  curved  line 
which  we  call  a  circle. 
Thus,  a  circle  is  a 
closed  curve,  all  points 
of  which  lie  in  the 
same  plane  and  are 
equally  distant  from  a 
fixed  point.  FIG.  38.  THE  CIRCLE 


52  GENERAL  MATHEMATICS 

65.  Center ;    circumference.    The  fixed  point    0  is  the 
center  of  the  circle.    The  length  of  the  curve  (circle)  is 
called  the  circumference  (distance  around)  of  the  circle. 

66.  Radius ;  diameter.    A  line  drawn  from  the   center 
of  a  circle  to  any  point  on  the  circle  is  a  radius.    Thus, 
OP  is  a  radius  of  the  circle  in  Fig.  38.    A  line  connecting 
two  points  on  the  circle  and  passing  through  the  center  of 
the  circle  is  called  a  diameter. 

From  the  definition  of  "  radius  "  given  above  it  is  clear 
that  in  a  given  circle  or  in  equal  circles  one  radius  has 
the  same  length  as  any  other.  Thus  we  obtain  the  follow- 
ing important  geometric  relation,  Radii  of  the  same  circle 
or  of  equal  circles  are  equal.  ("Radii"  is  the  plural  of 
"  radius.") 

67.  Arc;  to  intercept;  central  angle.    An  arc  is  a  part 
of  a  circle.    If  two  radii  are  drawn  from  the  center  of  the 
circle  to  two  different  points  on  the  circle,  they  cut  off  an 
arc  on  the  circle.   The  symbol  for."  arc  "  is  ""^    Thus,  AB  is 
read  "  the  arc  AB"   The  angle  formed  at  the  center  of  the 
circle  is  said  to  intercept  the  arc.    The  angle  at  the  center 
is  called  a  central  angle. 

68.  Quadrant;  semicircle.    An  arc  equal  to  one  fourth 
of  a  circle  is  called  a  quadrant.    An  arc  equal  to  one  half 
of  a  circle  is  called  a  semicircle. 

EXERCISES 

1 .  How  does  a  diameter  compare  in  length  with  a  radius  ? 

2.  How  many  quadrants  in  a  semicircle  ?  in  a  circle  ?    In 
what  connection  have  we  mentioned  the  idea  expressed  by  the 
word  "  quadrant "  ? 


PROPERTIES  OF  ANGLES 


53 


69.  Degrees  of  latitude  and  longitude.  The  use  that  is 
made  of  the  circle  in  geography  is  no  doubt  familiar  to 
all  of  us.  The  prime  meridian,  that  passes  through  Green- 
wich, England  (see  Fig.  39),  is  the  zero  meridian.  We 
speak  of  places  lying  to  the  west  of  Greenwich  as  being 
in  west  longitude  and  of  those  lying  to  the  east  of 
Greenwich  as  being  in  east  longitude  (see  Fig.  39).  Since 
it  takes  the  earth  twenty-four  hours  to  make  one  complete 
rotation,  the  sun  apparently  passes  over  one  twenty-fourth 


N.P. 


N.P. 


S.P. 
LATITUDE  AND  LONGITUDE 

of  the  entire  distance  around  the  earth  every  hour.  Thus, 
points  lying  a  distance  of  one  twenty -fourth  of  a  complete 
turn  apart  differ  in  time  by  an  hour. 

In  order  to  carry  the  computations  further  the  entire 
circle  is  divided  into  three  hundred  and  sixty  equal  parts, 
each  of  which  is  called  a  degree  (1°)  of  longitude.  In 
order  to  express  fractional  parts  of  the  unit  each  degree 
is  divided  into  sixty  minutes  (60')  and  each  minute  into 
sixty  seconds  (60").  With  this  agreement  the  longitude 
of  a  place  is  determined. 

The  position  of  a  place  north  or  south  of  the  equator  is  in- 
dicated by  the  number  of  degrees  of  north  or  south  latitude. 


54  GENERAL  MATHEMATICS 

EXERCISES 

1.  What  is  the  greatest  longitude  a  place  can  have  ?  the 
greatest  latitude  ? 

2.  How  many  seconds  are  there  in  a  degree  of  longitude  ? 
in  a  degree  of  latitude  ?  •  * 
(  $7  What   is    the   length   of   a   degree   arc    on  the   earth's 
surface  ?    of  a  minute  arc  ?    of  a  second  arc  ?    Try  to  find  out 
how  accurately  the  officers  of  a  ship  know  the  location  of 
the  ship  out  in  mid-ocean. 

4.  How  would  you  read  25°  14'  west  longitude  ?   33°  5'  17" 
north  latitude  ? 

5.  Compare  the  method  of  locating  by  latitude  and  longitude 
to  the  method  used  in  locating  a  house  in  a  large  city. 

6.  Find  out  in  what  longitude  you  live  ?  in  what  latitude  ? 

70.  Amount  of  rotation  determines  the  size  of  an  angle. 
If  we  remember  that  an  angle  is  formed  by  rotating  a  line 
in  a  plane  about  a  fixed  point,  it  will  be  clear  that  the 


Fio.  40 

size  of  the  angle  depends  only  on  the  amount  of  turning, 
not  upon  the  length  of  the  sides.  Since  the  sides  may 
be  extended  indefinitely,  an  angle  may  -have  short  or  long 
sides.  In  Fig.  40  the  angle  A  is  greater  than  angle  B,  but 
the  sides  of  angle  B  are  longer  than  the  sides  of  angle  A. 

71.  Measurement  of  angles;  the  protractor.  In  many 
instances  the  process  of  measuring  angles  is  as  important 
as  that  of  measuring  distances.  An  angle  is  measured 
when  we  find  how  many  times  it  contains  another  angle 
selected  as  a  unit  of  measure. 


PROPERTIES  OF  ANGLES 


55 


FIG.  41.   THE  PROTRACTOR 


The  protractor  (Fig.  41)  is  an  instrument  devised  for 
measuring  and  constructing  angles.  The  protractor  com- 
monly consists  of  a  semicircle  divided  into  one  hundred 
and  eighty  equal  parts.  Each  of  these  equal  parts  is  called 
a  degree  of  arc  (1°).  In  the  geography  work  referred  to  in 
Art.  69,  the  unit  for  longitude  and  latitude  was  the  degree 
of  arc.  In  the  measure- 
ment of  angles  we  shall 
consider  a  unit  corre- 
sponding to  a  unit  of 
arc  and  called  a  degree 
of  angle. 

If  straight  lines  are 
drawn  from  each  of 
these  points  of  division 
on  the  semicircle  to  the 
center  0,  one  hundred 
and  eighty  equal  angles  are  formed,  each  of  which  is  a 
degree  of  angle  (I0)-.  Thus,  the  unit  of  angular  measure 
is  the  degree.  A  degree  is  divided  into  sixty  equal  parts, 
each  of  which  is  called  a  minute  (!'). 

Each  minute  is  divided  into  sixty  equal  parts,  eack 
of  which  is  called  a  second  (1").  Of  course  the  minute 
and  the  second  graduations  are  not  shown  on  the  pro- 
tractor. Why  not  ? 

EXERCISES 

1 .  How  many  degrees  in  a  right  angle  ?  in  a  straight  angle  ? 
in  a  perigon  ? 

2.  A  degree  is  what  part  of  a  right  angle  ?    of  a  straight 
angle  ?   of  a  perigon  ? 

3.  What  angle  is  formed  by  the  hands  of  a  clock  at  3  o'clock  ? 
at  6  o'clock  ?   at  9  o'clock  ?  at  12  o'clock  ?  at  4  o'clock  ?  at 
7  o'clock  ?  at  11  o'clock  ? 


56  GENERAL  MATHEMATICS 

4.  Give  a  time  of  day  when  the  hands  of  a  clock  form  a 
right  angle;  a  straight  angle. 

5.  What  is  the  correct  way  to  read  the  following  angles : 
15°  17'  2"?   5°  0'  10"? 

6.  How  many  degrees  are  there  in  three  right  angles  ?    in 
four  straight  angles  ?    in  one  third  of  a  right  angle  ?  in  two 
thirds  of  a  straight  angle  ?  in  one  fifth  of  a  right  angle  ?   in 
x  straight  angles  ?   in  y  right  angles  ?  in  2  x  right  angles  ? 

7.  Ordinary  scales  for  weighing  small  objects  are  sometimes 
made  with  a  circular  face  like  a  clock  face.    The  divisions  of 
the  scale  indicate  pounds.    If  the  entire  face  represents  12  lb., 
what  is  the  angle  between  two  successive  pound  marks  ? 

8.  What  is  the  angle  between  two  successive  ounce  marks 
on  the  face  of  the  scale  in  Ex.  7  ? 

72.  Measuring  angles ;  drawing  angles.  The  protractor 
may  be  used  to  measure  a  given  angle.  Thus,  to  measure 
a  given  angle  x  place  the  protractor  so  that  the  center  of 
the  protractor  (point  0  in  Fig.  42)  falls  upon  the  vertex 
and  make  the  straight  edge  of  the  protractor  coincide 
with  (fall  upon)  the  initial  side  of  the  given  angle  x.  Now, 
observe  where  the  terminal  side  of  the  given  angle  intersects 
(crosses)  the  rim  of  the  protractor.  Read  the  number  of 
degrees  in  the  angle  from  the  scale  on  the  protractor. 

EXERCISES 

1.  Draw  three  different  angles,  one  acute,  one  obtuse,  and 
one  reflex.   Before  measuring,  estimate  the  number  of  degrees  in 
each  angle.   Find  the  number  of  degrees  in  each  angle  by  means 
of  the  protractor.    Compare  the  results  with  the  estimates. 

2.  Draw  freehand  an  angle  of  30°;  of  45°;  of  60°;  of  90°; 
of  180° ;  of  204°.    Test  the  accuracy  of  the  first  four  angles  by 
means  of  the  protractor. 


PEOPERTIES  OF  ANGLES 


57 


The  protractor  is  also  useful  in  constructing  angles 
of  a  required  size ;  for  example,  to  construct  an  angle 
of  42°  draw 
a  straight  line 
OX  (Fig.  42) 
and  place  the 
straight  edge 
of  the  pro- 
tractor on  the 

FIG.  42.    MEASURING  ANGLES  WITH  A  PROTRACTOR 
that  the  center 

rests  at  0.  Count  42°  from  the  point  on  OX  where  the 
curved  edge  touches  OX  and  mark  the  point  A.  Connect 
A  and  0,  and  the  angle  thus  formed  will  contain  42°. 

EXERCISES 

1.  Construct  an  angle  of  25°;  of  37°;  of  95°;  of  68°;  of 
112° ;  of  170°.    Continue  this  exercise  until  you  are  convinced 
that  you  can  draw  any  required  angle. 

2.  Construct  an  angle  equal  to  a  given  angle  ABC  by  means 
of  the  protractor. 

HINT.  Draw  a  working  line  MN.  Measure  the  angle  ABC. 
Choose  a  point  P  on  the  line  MN  as  a  vertex  and  construct  an 
angle  at  P  containing  the  same  number  of  degrees  as  the  given 
angle  ABC.  The  angle  is  then  constructed  as  required. 

3.  Draw  a  triangle.    How  many  angles  does  it  contain  ? 
Measure  each  with  the  protractor. 

73.  How  to  measure  angles  out  of  doors.  It  is  possible 
to  measure  angles  out  of  doors  by  means  of  a  simple  field 
(out-of-door)  protractor,  so  that  some  simple  problems  in  sur- 
veying can  be  solved.  Such  a  field  protractor  may  be  made 
by  a  member  of  the  class,  as  shown  on  the  following  page. 


58 


GENERAL  MATHEMATICS 


Secure  as  large  a  protractor  as  possible  and  fasten  it  on 
an  ordinary  drawing  board.  Attach  the  board  to  a  camera 
tripod  (if  this  is  not  to  be  had,  a  rough  tripod  can  be 
made).  Make  a  slender  pointer  which  may  be  attached  at 
the  center  of  the  circle  with  a  pin  so  that  it  may  swing  freely 
about  the  pin  as  a  pivot.  Place  two  inexpensive  carpenter's 
levels  on  the  board,  and  the  instrument  is  ready  for  use. 

Thus,  to  measure  an  angle  ABC  (suppose  it  to  be  an 
angle  formed  by  the  intersection  of  an  avenue,  BA,  and  a 
street,  -BC),  first  put  the  board  in  a  horizontal  position 
(make  it  stand  level).  Then  place  the  center  of  the  circle 
over  the  vertex  of  ; 

the  angle  to  be  meas- 
ured and  sight  in 
the  direction  of  each 
side  of  the  angle, 
noting  carefully  the 
reading  on  the  pro- 
tractor. The  number 
of  degrees  through 
which  the  pointer 
is  turned  in  passing 
from  the  position  of 
BA  to  that  of  BC 
is  the  measure  of 
angle  ABC. 

74.  Transit.  When 
it  is  important  to 
secure  a  higher  de- 
gree of  accuracy  than  is  possible  with  the  instrument 
described  in  Art.  73,  we  use  an  instrument  called  a  transit 
(Fig.  43).  This  instrument  is  necessary  in  surveying. 
Three  essential  parts  of  the  transit  are  (1)  a  horizontal 


FIG.  43.    THE  TRANSIT 


59 

graduated  circle  for  measuring  angles  in  the  horizontal 
plane  (see  D  in  Fig.  43)  ;  (2)  a  graduated  circle,  C,  for 
measuring  angles  in  the  vertical  (up-and-down)  plane  ;  and 
(3)  a  telescope,  AB,  for  sighting  in  the  direction  of  the 
sides  of  the  angle.  For  a  fuller  description  of  the  transit 
see  a  textbook  in  trigonometry  or  surveying. 

HISTORICAL  NOTE.  The  division  of  the  circle  into  three  hundred 
and  sixty  degrees  and  each  degree  into  sixty  minutes  and  each 
minute  into  sixty  seconds  is  due  to  the  Babylonians.  Cajori  cites 
Cantor  and  others  somewhat  as  follows :  At  first  the  Babylonians 
reckoned  the  year  as  three  hundred  and  sixty  days.  This  led  them 
to  divide  the  circle  into  three  hundred  and  sixty  degrees,  each  degree 
representing  the  daily  part  of  the  supposed  yearly  revolution  of  the 
sun  around  the  earth.  Probably  they  were  familiar  with  the  fact  that 
the  radius  could  be  applied  to  the  circle  exactly  six  times  and  that 
as  a  result  each  arc  cut  off  contained  sixty  degrees,  and  in  this  way 
the  division  into  sixty  equal  parts  may  have  been  suggested.  The 
division  of  the  degree  into  sixty  equal  parts  called  minutes  may  have 
been  the  natural  result  of  a  necessity  for  greater  precision.  Thus  the 
sexagesimal  system  may  have  originated.  "  The  Babylonian  sign  *  is 
believed  to  be  associated  with  the  division  of  the  circle  into  six  equal 
parts,"  and  that  this  division  was  known  to  the  Babylonians  seems 
certain  "  from  the  inspection  of  the  six  spokes  in  the  wheel  of  a  royal 
carriage  represented  in  a  drawing  found  in  the  remains  of  Nineveh." 

Henry  Briggs  attempted  to  reform  the  system  by  dividing  the 
degree  into  one  hundred  minutes  instead  of  into  sixty,  and  although 
the  inventors  of  the  metric  system  are  said  to  have  proposed  the 
division  of  the  right  angle  into  one  hundred  equal  parts  and  to 
subdivide  decimally,  instead  of  the  division  into  ninety  parts,  we 
have  actually  clung  to  the  old  system.  However,  there  is  a  tend- 
ency among  writers  to  divide  each  minute  decimally ;  for  example, 
52°  10.2'  instead  of  52°  10'  12".  See  Cajori,  "  History  of  Elementary 
Mathematics,"  1917  Edition,  pp.  10,  43,  and  163. 

75.  Comparison  of  angles.  In  order  to  make  a  comparison 
between  two  angles,  we  place  one  over  the  other  so  that  the 
vertex  and  the  initial  side  of  one  coincide  with  the  vertex  and 
the  initial  side  of  the  other.  If  the  terminal  sides  coincide, 


60 


GENERAL  MATHEMATICS 


the  angles  are  equal ;  if  the  terminal  sides  do  not  coincide, 
the  angles  are  unequal — assuming,  of  course,  in  both  cases, 
that  each  of  the  two  angles  compared  is  less  than  360°.  In 
the  exercises  and  articles  that  follow  we  consider  no  angle 
greater  than  360°. 

EXERCISES 

1.  Compare  angles  x,  y,  and  2,  in  Fig.  44,  and  arrange  them 
in  order  as  to  size. 

HIXT.  Make  a  tracing  of  each  on  thin  paper  and  try  to  fit  each 
on  the  other. 


FIG.  44 

2.  Construct  an  angle  equal  to  a  given  angle  ABC.  Lay  a  thin 
sheet  of  paper  over  the  angle  ABC  and  make  a  tracing  of  it.   Cut 
out  the  tracing  and  paste  it  to  another  part  of  the  paper.    The 
angle  thus  shown  is  equal  to  the  angle  ABC. 

3.  Try  to  draw  freehand  two  equal  angles. 
Test  your  drawings  by  the  method  of  Ex.  1. 

4.  Draw    freehand    one    angle    twice    as 
large  as  another.    Test  your  drawings  with 
the  protractor. 


FIG.  45 


76.  Adjacent  angles ;  exterior  sides. 
Angles  x  and  y  in  Fig.  45  are  two  angles 
which  have  a  common  vertex  and  a  com- 
mon side  between  them.  The  angles  x  and  y  are  said  to  be 
adjacent  angles.  Thus,  adjacent  angles  are  angles  that  have 
the  same  vertex  and  have  a  common  side  between  them. 
The  sides  OT  and  OR  are  called  the  exterior  sides. 


PKOPERTIES  OF  ANGLES 


61 


E 


EXERCISES 

1.  Indicate  which  angles  in  Fig.  46  are  adjacent.   Point  out 
the  common  vertex  and   the  common  side  in  each  pair  of 
adjacent  angles. 

2.  Draw  an  angle  of  45°  adjacent 
to  an  angle  of  45°;  an  angle  of  30° 
adjacent  to  an  angle   of  150°;   an 
angle  of  35°  adjacent   to   an   angle 
of  80°. 

3.  Do  you  notice  anything  partic- 
ularly significant  in  any  of  the  parts 
of  Ex.  2  ? 

4.  Draw  an  angle  of  30°  adjacent 
to  an  angle  of  60°.    What  seems  to 
be  the  relation  between  their  exterior 
sides  ?    Does   this    relation  need  to 
exist  in  order  that  the  angles  shall 

be  adjacent  ?   What  total  amount  of  turning  is  represented  ? 

77.  Geometric  addition  and  subtraction  of  angles.  Exs.  2 

and  4,  above,  suggest  a  method  for  adding  any  two  given 
angles.   Thus,  to  add  a  given  angle  y  to  a  given  angle  x, 

B 


FIG.  46 


FIG.  47.    GEOMETRIC  ADDITION  OF  ANGLES 

Fig.  47,  angle  y  is  placed  adjacent  to  angle  z,  and  the  re- 
sulting angle  is  called  the  sum  of  x  and  y.  The  angles  may 
be  transferred  to  the  new  position  either  by  means  of  tracing 
paper  or,  more  conveniently,  by  means  of  the  protractor. 


62  GENERAL  MATHEMATICS 

EXERCISE 

Add  two  angles  by  placing  them  adjacent  to  each  other. 

We  may  also  find  the  difference  between  two  angles. 
In  Fig.  48  the  two  given  angles  are  x  and  y.  Place  tin- 
smaller  angle,  y,  on  the  larger,  a:,  so  that  the  vertices  and 


o 


FIG.  48.    GEOMETRIC  SUBTRACTION  OF  ANGLES 

one  pair  of  sides  coincide.  The  part  remaining  between  the 
other  two  sides  of  x  and  y  will  be  the  difference  between 
x  and  y.  Thus,  in  Fig.  48  we  obtain  Zz  —  /.y=/.AOC. 

EXERCISES 

1.  Draw  three  unequal  angles  x,  y,  and  2,  so  that  y~>x  and 
x  >  z.    Draw  an  angle  equal  to  x  +  y  +  z  ;  equal  to  y  —  x  +  z  ; 
equal  to  y  -\-  x  — -  z. 

2.  Draw  an  angle  of  60°  and  draw  another  of  20°  adjacent 
to  it.    What  is  their  sum  ?    Fold  the  20-degree  angle  over  the 
60-degree  angle  (subtraction)  and  call  the  difference  x.    What 
is  the  equation  which  gives  the  value  of  x  ? 

78.  Construction  problem.  At  a  given  point  on  a  given 
line  to  construct  by  means  of  ruler  and  compasses  an 
angle  equal  to  a  given  angle.  In  this  construction  we 
make  use  of  the  following  simple  geometric  relation  be- 
tween central  angles  and  their  intercepted  arcs :  In  the 
same  circle  or  in  equal  circles  equal  central  angles  intercept 


PROPERTIES  OF  ANGLES  63 

equal  arcs  on  the  circle.  For  example,  if  the  central  angle 
contains  nineteen  angle  degrees,  then  the  intercepted  arc 
contains  nineteen  arc  degrees. 

The  student  may  possibly  see  that  this  geometric  rela- 
tion is  implied  in  our  definitions  of  Art.  71.  However, 
the  two  following  paragraphs  will  assist  him  in  under- 
standing its  application. 

Make  a  tracing  of  the  circle  and  the  angle  ABC  .in 
Fig.  49,  (a),  and  place  B  upon  E  in  Fig.  49,  (b).  The 
angles  must  coincide  because  they  are  given  equal.  Then 
the  circle  whose  center 
is  B  (circle  B)  must 
coincide  with  the  cir- 
cle whose  center  is  E 
(circle  -E1),  because  the 
radii  of  equal  circles 
are  equal.  Then  A  will 

FIG.  49 
fall  on  Z>,  and  C  on  F; 

that  is,  the  arc  CA  will  fall  on  the  arc  FD,  and  the  arcs 
are  therefore  equal. 

It  is  easy  to  see  that  the  following  statement  is  also 
true :  In  the  same  circle  or  in  equal  circles  equal  arcs  on  the 
circle  are  intercepted  by  equal  central  angles.  For  circle  M 
can  be  placed  on  circle  E  so  that  arc  CA  coincides  with 
arc  FD,  since  these  arcs  are  given  equal,  and  ''so  .that  B 
falls  on  E.  A  will  fall  on  Z>,  and  C  on  F.  Then  the  angles 
must. coincide  and  are  therefore  equal. 

The  two  preceding  geometric  relations  make  clear  why 
the  protractor  may  be  used  to  measure  angles  as  we  did 
in  Art.  71.  The  method  used  there  is  based  upon  the  idea 
that  every  central  angle  of  one  degree  intercepts  an  arc 
of  one  degree  on  the  rim  of  the  protractor ;  that  is,  when 
we  know  the  number  of  degrees  in  an  angle  at  the  center 


64 


GENEKAL  MATHEMATICS 


of  a  circle  we  know  the  number  of  degrees  in  the  arc 
intercepted  by  its  sides,  and  vice  versa. 

The  idea  can  be  expressed  thus:  A  central  angle  is 
measured  by  the  arc  intercepted  by  its  sides  (when  angular 
degrees  and  arc  degrees  are  used  as  the  respective  units 
of  measure). 

How  many  degrees  of  an  arc  are  intercepted  by  a  central 
angle  of  30°  ?  of  40°  ?  of  60.5°  ?  of  n°  ? 

We  are  now  ready  to  proceed  with  our  problem:  At 
a  given  point  on  a  given  line  to  construct  by  means  of 
ruler  and  compasses  an  angle  equal  to  a  given  angle. 


FIG.  60.   CONSTRUCTING  AN  ANGLE  EQUAL  TO  A  GIVEN  ANGLE 


Construction.  Let  DEF  in  Fig.  50  be  the  given  angle  and  let  P 
be  the  given  point  on  the  given  line  AB. 

With  E  as  a  center  and  ER  as  a  radius  draw  a  circle.  With  P  as 
a  center  and  with  the  same  radius  {ER^  draw  another  circle.  Place 
the  sharp  point  of  the  compasses  at  R  and  cut  an  arc  through  M. 
With  S  as  a  center  and  the  same  radius  cut  an  arc  at  N. 

The  Z.BPC  is  the  required  angle.     Why? 


EXERCISES 


1.  Check  the  correctness  of  your  construction  for  the  pre- 
ceding directions  by  measuring  with  a  protractor. 

2.  How  many  ways  have  we  for  constructing  an  angle  equal 
to  a  given  angle  ? 


PROPERTIES  OF  ANGLES  65 

3.  Construct  two  angles,  designating  one  of  them  as  contain- 
ing a  degrees  and  the  other  as  containing  b  degrees ;  using  the 
angles  a  and  b,  construct  an  angle  containing  a  +  b  degrees ; 
construct  an  angle  containing  2  a  -f-  b  degrees. 

4.  Choose  a  and  b  in  Ex.  3  so  that  a  >b,  then  construct  an 
angle  equal  to  the  difference  of  the  two  given  angles. 

5.  Construct  an  angle  equal  to  the  sum  of  three  given  angles. 

79.  Perpendicular.    We  have  seen  in  Art.  61  how  right 
angles   are  formed  by  rotation.    If  two  lines  form  right 
angles  with  each  other,  they  are  said  to  be  perpendicular 
to  each  other.   The  symbol 

for  "  perpendicular  "  is  _L.  ^ 

80.  Construction  problem. 
At  a  given  point  on  a  given 
line  to  erect  a  perpendic- 
ular to  that  line  by  using    A 

ruler  and  compasses.  c\  ID 

FlG.  51.      HOW   TO    ERECT   A 

Construction.   Let  AB  be  the  PERPENDICULAR 

given   line    and   P   the   given 

point  (Fig.  51).  With  P  as  a  center  and  with  a  convenient  radius 
draw  arcs  intersecting  AB~&t  C  and  D. 

With  C  and  D  as  centers  and  with  a  radius  greater  than  \  CD 
draw  arcs  intersecting  at  E.  Draw  EP.  Then  EP^s  the  required 
perpendicular. 

EXERCISES 

1.  Test  the  accuracy  of  your  construction  in  the  problem  of 
Art.  80  by  using  a  protractor. 

2.  Why  must  the  radius  CE  in  Fig.  51  be  greater  than  ^  CD? 

3.  In  Fig.  52  how    — < =- 

A 
would    you    draw    a  ^IG  §2 

perpendicular   to    the 

line  AB  at  the  point  A?    Do  the  construction  work  on  paper. 


06  GENERAL  MATHEMATICS 

81.  Construction   problem.    How  to  bisect  a  given  line 
segment  AB. 

Construction.  Let  AB  be  the  given 
line  segment  (Fig.  53).  With  A  as 
a  center  and  with  a  radius  greater 

A  I  I  171 

than  ^  AB  describe  arcs  above  and    * 
below  AB.    With  B  as  a  center  and 
with  the  same  radius  as  before  de- 
scribe   arcs    above    and    below   and  \  /~ 
intersecting  the  first  arcs  at  C  and  D. 
Draw  CD.    Then  E  is  the  point  of        FIG.  53.    How  TO  BISECT 
bisection  for  AB.                                                  *  LINE  SEGMENT 

82.  Perpendicular  bisector.    The  line   CD  in.  Fig.  53  is 
called  the  perpendicular  bisector  of  AB. 

EXERCISES 

1.  How  may  a  line  be  divided  into  four  equal  parts  ?  into 
eight  equal  parts  ? 

2.  Draw  a  triangle  all  of  whose  angles  are  acute  (acute- 
angled   triangle).     Construct   the   perpendicular   bisectors    of 
each  of  the  three  sides  of  the  triangle. 

3.  Cut  out  a  paper  triangle  and  fold  it  so  as   to  bisect 
each  side. 

4.  Draw  a  trianglfe  in  which  one  angle  is  obtuse  (obtuse- 
angled  triangle)  and  draw  the  perpendicular  bisectors  of  the 
three  sides. 

5.  Draw  a  triangle  in  which  one  angle  is  a  right  angle  and 
construct  the  perpendicular  bisectors  of  the  sides. 

6.  Draw  a  triangle  ABC.    Bisect  each  side  and  connect  each 
point  of  bisection  with  the  opposite  vertex. 

83.  Median.    A  line  joining  the  vertex  of  a  triangle  to 
the  mid-point  of  the  opposite  side  is  called  a  median. 


PKOPEKTIES  OF  ANGLES 


EXERCISE 
Draw  a  triangle  ;  construct  its  medians. 

84.  Construction    problem.    From    a    given    point    out- 
side a  given  line   to  drop   a  perpendicular  to  that  line. 

Construction.  Let  AB  be  the  given  line  and  P  the  given  point 
(Fig.  54).    With  P  as  a  center  and  p 

with  a  radius  greater  than  the  dis- 
tance from  P  to  AB  describe  an  arc 
cutting  AB  at  M  and  R.  With  M 
and  R  as  centers  and  with  a  radius 
greater  than  \  MR  describe  arcs 
either  above  or  below  (preferably  be- 
low) the  line  AB.  Connect  the  point 
of  intersection  E  with  P.  Then  the 
line  PD  is  perpendicular  to  AB,  as 
required.  Test  the  accuracy  of  your 
work  by  measuring  an  angle  at  D. 


\/E 


FIG.  54.    How  TO  DROP  A. 
PERPENDICULAR 


EXERCISES 

1  .  Why  is  it  preferable  to  describe  the  arcs  in  Fig.  54  below 
the  line  AB  ? 

2.  Draw  a  triangle  ABC  all  of  whose  angles  are  acute  and 
draw  perpendiculars  from  each  vertex  to  the  opposite  sides. 

85.  Altitude.  An  altitude  of  a  triangle  is  a  line  drawn 
from  a  vertex  perpendicular  to  the  opposite  side. 


EXERCISES 

1.  Draw  a  triangle  in  which  one  angle  is  obtuse  and  draw 
the  three  altitudes. 

2.  Draw  a  triangle  in  which  one  angle  is  a  right  angle  and 
draw  the  three  altitudes. 

3.  When  d.o  the  altitudes  fall  inside  a  triangle  ?   outside? 


68 


GENERAL  MATHEMATICS 


86.  To  bisect  a  given  angle.  Suppose  angle  ABC  to 
be  the  given  angle  (Fig.  55).  With  the  vertex  B  as  a 
center  and  with  a  convenient  radius  draw  an  arc  cutting 
HA  and  BC  at  X  and  Y  re- 
spectively. With  X  and  1' 
as  centers  and  with  a  radius 
greater  than  ^  XY  draw  arcs 
meeting  at  D.  Join  B  and  D. 
Then  BD  is  the  bisector  of 
/.ABC. 


FIG.  55.    How  TO  BISECT 
AN  ANGLK 


EXERCISES 

1.  Bisect  an  angle  and  check  by  folding  the  paper  so  that 
the  crease  will  bisect  the  angle. 

2.  Bisect  an  angle  of  30°  ;  of  45° ;  of  60°  ;  of  5)0°. 

3.  Divide  a  given  angle  into  four  equal  parts. 

4.  Draw  a  triangle  whose  angles  are  all  acute  and  bisect 
each  of  the  angles. 

5.  Draw  a  triangle  in  which  one  angle  is  obtuse  and  bisect 
each  of  the  angles ;  do  the  same  for  a  triangle  in  which  one 
angle  is  a  right  angle. 

87.  Parallel  lines.  AB  and  CD  in  Fig.  5(3  have  had 
the  same  amount  of  angular  rotation  from  the  initial  line 
EF.  Thus,  they  have  the 
same  direction  and  are  said 
to  be  parallel.  The  symbol 
for  ''parallel"  is  II.  Thus, 
AB  II  CD  is  read  "AB  is  par- 
allel to  CD" 


A  C 

FIG.  56.    PARALLEL  LINES 


88.  Corresponding  angles ; 
transversal.  Angles  x  and 
//  in  Fig.  56  are  called  corresponding  angles.  The  line  EF 


69 

is  called  a  transversal.  It  is  clear  that  the  lines  are  par- 
allel only  when  the  corresponding  angles  are  equal  and 
that  the  corresponding  angles  are  equal  only  when  the 
lines  are  parallel. 

EXERCISES 

1.  Draw  figures   to   illustrate   the   importance  of   the  last 
statement  in  Art.  88,  above. 

2.  Point  out  the  parallel  lines  you  can  find  in  the  classroom. 

89.   Construction  problem.    How  to  draw  a  line  parallel 
to  a  given  line. 

Construction.    Choose    a    point   P 
outside  the  given  line  A  B  in  Fig.  57. 

Draw  a  line  through  P  so  as  to  form     

a  convenient  angle  x  with  AR.    Call 

the  point  of  intersection  D.    At  P, 

using  DP  as  initial  line,  construct  an 

angle  y  equal  to  angle  x  (as  shown) 

by  the  method  of  Art.  78.    Then  PR    j[ 

and  AB   are    parallel    because  they 

have  had  the  same  amount  of  rota-          FIG.  57.    How  TO  DKAW 

tion  from  the  initial  line  PD.  PARALLEL  LISKS 

EXERCISES 

1.  Construct  a  line  parallel  to  a  given  line  through  a  given 
point  outside  the  line. 

2.  A  carpenter  wants  a  straight-edge  board  to  have  parallel 
ends.    He  makes  a  mark  across  each  end  with  his  square.    Why 
will  the  ends  be  parallel  ? 

3.  In  Fig.  57  if  angle  x  =  60°,  what  is  the  number  of  degrees 
in  Z.  y  ?    Give  a  reason  for  your  answer. 

4.  Two  parallel  lines  are  cut  by  a  transversal  so  as  to  form 
two  corresponding  angles  (x  + 125°)  and  (3  x  +  50°).  Find  x  and 
the  size  of  each  angle.    Make  a  drawing  to  illustrate  your  work. 


GENERAL  MATHEMATICS 


FIG.  58 


5.  In  Fig.  58  if  AB  II  CD,  what  other  angles  besides  x  and  y 
are  equal  corresponding  angles  ? 

6.  In  Fig.  58,  /.x  =  Z.y.   Bisect 
Z  x  and  Z  y  and  show  that  these 
bisectors  are  parallel  to  each  other. 

90.  Parallelogram.  If  onepair 
of  parallel  lines  cross  (intersect) 
another    pair,     the     four-sided 
figure  thus  formed  is  called  a 
parallelogram ;  that  is,  a  paral- 
lelogram is  a  quadrilateral  whose  opposite  sides  are  parallel. 

91.  How  to  construct  a  parallelogram.    If  we  remember 
the  method  used  in  Art.  89  for  constructing  one  line  parallel 
to  another,  it  will  be 

easy  to  construct  a 
parallelogram.  Thus, 
draw  a  working  line 
AB  (Fig.  59).  Draw 
AR  making  a  conven- 
ient angle  with  AB. 
Through  any  point,  as 
P,  on  AR  draw  a  line 
P  V parallel  to  AB.  Through  any  point  M on  A B  draw  a  line 
MT  parallel  to  AR.  The  figure  AMSP  is  a  parallelogram, 
for  its  opposite  sides  are  parallel. 

92.  Rectangle.    If    one    of   the  in- 
terior angles  of  a  parallelogram  is  a 
right  angle,  the  figure  is  a  rectangle 
(Fig.  60).     Thus,    a    rectangle    is    a 

parallelogram  in  which  one  interior  angle  is  a  right  angle. 


FIG.  59.    How  TO  CONSTRUCT  A 
PARALLELOGRAM 


FIG.  60 


PROPERTIES  OF  ANGLES  71 

EXERCISE 

Show  that  all  the  angles  of  a  rectangle  are  right  angles. 
HINT.    Extend  the  sides  of  the  rectangle. 

93.  Square.    If   all   the   sides  of   a  rec- 
tangle   are    equal,    the    figure    is    called   a 

square  (Fig.  61). 

FIG.  61 
EXERCISES 

1.  Give  examples  of  rectangles  ;  of  squares. 

2.  Construct  a  rectangle  having  two  adjacent  sides  equal  to 
5  cm.  and   8  cm.   respectively    (use    compasses  '  and   straight- 
edge only).  ,  a  , 

3.  Construct  a  rectangle  hav-  ,  . 
ing  the  two  adjacent  sides  equal 

to  the  line  segments  a  and  b  in  FlG-  62 

K-  62.. 


- 

4.  Construct  a  square  whose 

side  is  7  cm.  long.  FlG>  63 

5.  Construct  a  square  a  side  of  which  is  a  units  long  (use 
line  a  in  Fig.  63). 

SUMMARY 

94.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases  :  angle,  vertex,  vertices,  initial  side 
of  an  angle,  terminal  side  of  an  angle,  right  angle,  straight 
angle,  perigon,  acute  angle,  obtuse  angle,  reflex  angle, 
circle,  center,  circumference,  radius,  diameter,  radii,  arc, 
intercept,  central  angle,  quadrant,  semicircle,  latitude,  lon- 
gitude, degree  of  latitude,  degree  of  longitude,  minute, 
second,  size  of  an  angle,  protractor,  degree  of,  arc,  degree 
of  angle,  adjacent  angles,  exterior  sides  of  an  angle,  field 


7^  (JKNERAL  MATHEMATICS 

protractor,  transit,  perpendicular  bisector,  perpendicular  to 
a  line,  altitude  of  a  triangle,  median  of  a  triangle,  bisector 
of  an  angle,  parallel  lines,  corresponding  angles,  transversal, 
parallelogram,  rectangle,  and  square. 

95.  The  following  symbols  have  been  introduced:  Z  for 
angle ;   rt.  /.   for   right   angle ;   A  for   angles ;    •   v  for   arc ; 
_L  for  is  perpendicular  to ;   II  for  is  parallel  to ;  °  for  dei/n-f 
or  degrees  ;   '  for  minute  or  minutes ;   "  for  second  or  seconds. 

96.  The  following  notations  have  been  discussed :  (1)  no- 
tation for  denoting  and  reading  angles;  (2)  notation  for 
denoting  a  circle  by  its  center. 

97.  This  chapter  has  presented  the  important  methods  of 

1.  Classifying  angles. 
'  2.  Measuring  angles. 

3.  Comparing  angles. 

4.  Drawing  angles  containing  any  amount  of  turning  or 
any  number  of  degrees. 

5.  Adding  and  subtracting  angles. 

6.  Measuring  angles  out  of  doors. 

98.  In  this  chapter  the  pupil  has  been  taught  the  follow- 
ing fundamental  constructions : 

1.  To  draw  a  circle. 

2.  To  draw  an  angle  equal  to  a  given  angle. 

3.  To  draw  a  line  perpendicular  to  a  given  line  at  a 
given  point. 

4.  To  draw  the  perpendicular  bisectors  of  the  sides  of 
a  triangle. 

5.  To  draw  the  medians  of  a  triangle. 

6.  To  draw  a  line  perpendicular  to  a  given  line  from  a 
given  point  outside  the  line. 

7.  To  draw  the  altitudes  of  a  triangle. 


PBOPEKTIES  OF  ANGLES  73 

8.  To  bisect  a  given  angle. 

9.  To  draw  the  bisectors  of  the  angles  of  a  triangle. 

10.  To  draw  a  line  through  a  given  point  parallel  to  a 
given  line. 

11.  To  construct  a  parallelogram. 

12.  To  construct  a  rectangle. 
18.  To  construct  a  square. 

99.  This  chapter  has  taught  the  pupil  to  use  the  follow- 
ing instruments  and  devices :  tracing  paper,  the  protractor, 
and  the  field  protractor. 

IMPORTANT  GEOMETRIC  RELATIONS 

100.  Radii  of  the  same  circle  or  of  equal  circles  are  equal. 

101.  In  the  same  circle  or  in  equal  circles  equal  central 
angles  intercept  equal  arcs  on  the  circle. 

102.  In  the  same  circle  or  in  equal  circles  equal  arcs  on 
the  circle  are  intercepted  by  equal  central  angles. 

103.  A  central  angle  is  measured  by  the  arc  intercepted 
by  its  sides. 

IMPORTANT  DEFINITIONS 

.  104.  An  angle  is  the  amount  of  turning  made  by  a  line 
rotating  about  a  fixed  point  in  a  plane. 

105.  A  circle  is  a  closed  curve  all  points  of  which  lie  in 
the  same  plane  and  are  equidistant  from  a  fixed  point. 

106.  (1)  A  quadrilateral  whose  opposite  sides  are  parallel 
is  a  parallelogram.     (2)   A  rectangle  is  a  parallelogram  in 
which  one  interior  angle  is  a  right  angle.    (3)  A  square  is  a 
rectangle  with  all  sides  equal. 


CHAPTER  IV 


FIG.  64. 


THE  EQUATION  APPLIED  TO  AREA 

107.  Measuring  areas.    If  we  determine  the  amount  of 
area  inclosed  within   a  polygon,  as  in  the  triangle  ABC 
in  Fig.  64,  we  are  measuring  the  area  of  the  triangle.    As  in 
measuring  length,  the  process  is 

one  of  comparison.   We  compare 

the  area  of  the  given  polygon 

with  some  standardized  (defined 

and  accepted)  unit  of  area  and 

determine  how  many  units  are  contained  in  the  polygon ; 

that  is,  we  determine  the  ratio  between  the  area  of  the 

given  polygon  and  a  standard  unit  of  area. 

108.  Unit  of  area.    The  unit  of  area  is  a  square  each 
of  whose  sides  is  a  standard  unit  of  length.    Such  a  unit 
involves  length  and  width.    Thus,  we  may 

measure  area  and  express  the  result  in 
square  feet,  square  inches,  square  meters, 
square  centimeters,  etc. 

109.  Practical  method  of  estimating  area. 
A  practical  way  to  estimate  the  area  of  a 
polygon  is  to  transfer  it  to  squared  paper 

by  means  of  tracing  paper  and  then  count  the  num- 
ber of  square  units  inclosed  within  the  figure.  If  the 
bounding  lines  cut  the  squares,  it  becomes  necessary  to 
approximate.  In  such  approximations  we  should  be  careful, 
but  we  should  not  go  beyond  reasonable  limits  of  accuracy. 

74 


1cm. 


1  cm. 


FIG.  65.    UNIT  OF 
AREA  ix  THE  MET- 
RIC SYSTEM 


THE  EQUATION  APPLIED  TO  AREA 


75 


EXERCISES 

1.  The  six  figures  in  Fig.  66  were  transferred  by  means  of 
tracing  paper.  Estimate  the  areas  of  each  of  them  by  counting 
the  squares.  Express  the  areas  either  as  square  centimeters 
or  as  square  millimeters. 

HINT.    One  small  square  equals  4  sq.  mm. 


FIG.  66.    ESTIMATING  AREAS  BY  MEANS  OF  SQUARED  PAPER 

2.  If  the  paper  were   ruled  much  finer,  would  you  get  a 
more  accurate  estimate  ?    Give  an  argument  for  your  answer. 

3.  Do  you  think  that  any  of  your  results  are  accurate? 

4.  Write    a    paragraph    in 
precise  terms,  supporting  your 
answer  to  Ex.  3. 


C- 


110.  Area  of  a  rectangle. 
CASE  I.  The  sides  of  the  rec- 
tangle that  has  been  trans- 
ferred to  the  squared  paper  of 
Fig.  67  are  integral  multiples 
of  1  cm.  Using  1  sq.  cm.  as  a  unit,  there  are  two  rows  of 


FIG.  67.    How  TO  FIND  THE  AREA 
OF  A  RECTANGLE 


76 


GENERAL  MATHEMATICS 


units,  and  four  units  in  a  row.  Counting,  we  see  that  the 
area  equals  8,  or  2  x  4.  The  law  in  this  case  is :  The  area 
equals  the  base  times  the  altitude.  In  equation  form  this 
law  may  be  written  A  =  b  x  a. 

111.  Area  of  a  rectangle.    CASE  II.   Let  us  suppose  that 
we  are  given  a  square  whose  sides  are  not  integral  multiples 
of  1  cm. :   for  example,   a  rectangle  whose  base  (length) 
is  2.3  cm.  and  whose  altitude  (width)  is  1.3  cm.    If  we 
assume  that  the  preceding  law  holds,  then  we  ought  to 

get  2.3  x  1.3  =  2.99  sq.  cm.    Instead     

of  putting  the  rectangle  on  the  kind 

of  squared  paper  used  in  Case  I, 

let   us    draw    it    again,    by    means 

of  tracing  paper,  on  squared  paper 

that    is    ruled    to    a   smaller   unit, 

the  millimeter,  as  in  Fig.  68.    Since 

there   are   23  mm.   in    2.3  cm.   and 

13  mm.  in   1.3  cm.,   if  we   temporarily  adopt  the   square 

millimeter  as  a  unit  of  area,  then  the  sides  of  the  rectangle 

are,  as  in  Case  I,  integral  multiples  of  the  unit  of  length 

(in  this  case  the  millimeter).     Hence  there  are   13  rows 

of  units  with  23  in  a  row,  or  299  sq.  mm.    But  there  are 

100  sq.  mm.  in  1  sq.  cm. ;  hence,  dividing  299  by  100,  the 

result  is  2.99  sq.  cm.,  which  is  precisely  the  same  number 

as  that  obtained  by  assuming  the  law  of  Case  I. 

112.  Area  of  a  rectangle.    CASE  III.    This  process  of 
temporarily  adopting  a  smaller  square  can  be  continued. 
If,  for  example,  the  base  of  a  rectangle  is  2.13  cm.  and 
the  altitude  1.46  cm.,  we  may  imagine  the  rectangle  to  be 
drawn  upon  squared  paper  still  finer  ruled,  that  is,  ruled 
to  0.1  of  a  millimeter.    From  here  the  reasoning  is  the 
same  as  in  Cases  I  and  II. 


FIG. 


THE  EQUATION  APPLIED  TO  AREA 


77 


EXERCISES 

1.  Finish  the  reasoning  of  the  foregoing  paragraph. 

2.  The  base  of  a  rectangle  is  3^  cm.  and  its  altitude  is  5-|  cm. 
Show  that  the  area  may  be  found  by  counting  squares. 

3.  The  base  of  a  rectangle  is  5|  cm.  and  its  altitude  is-2|  cm. 
What  unit  would  you  temporarily  adopt  to  find  the  area? 
Express  the  area  in  square  centimeters. 

The  preceding  exercises  show  that  if  the  sides  of  a 
rectangle  involve  fractions  which  may  be  expressed  as 
exact  decimal  parts  of  a  unit,  the  problem  is  the  same 
as  in  Case  II. 

113.  Second  method  for  finding  area  of  a  rectangle.    It  is 

possible  to  show  that  the  transfer  of  a  rectangle  to  the 

squared  paper  by  means  of  tracing  paper  was  unnecessary. 

Suppose  we  are  given  a  rec- 

tangle  AS  CD  (see  Fig.  69) 

whose    base    is    4  cm.    and 

whose  altitude  is  3  cm.  We 

wish  to  find  the  area.   Draw 

a  perpendicular  line  to  the 

line  AB  at  the  end  of  each 

unit  segment ;  that  is,  at  the 

points  E,  F,  and  G  (review 

the    method    of   Art.   80). 

Also  construct  perpendiculars  to  the  line  AD  at  the  points 

H  and  L    Then  each  small  square  is  a  unit  of  measure 

(by  definition),  and  the  figure  is  divided  into  three  rows 

of  units  with  four  in  a  row.    By  counting,  the  area  equals 

12  (that  is,  base  times  altitude'). 


II 
I 

A 

E           F           G           £ 
FIG.  69 

78  GENERAL  MATHEMATICS 

EXERCISES 

1.  The  base  of  a  rectangle  is  4.3cm.  long,  and  its  altitude 
is  1.7  cm.   Show  how  to  find  the  area  of  this  rectangle  by  count- 
ing, but  without  the  use  of  squared  paper. 

2.  Apply  the  law  A  =  b  x  «.    What  advantage  has  this  law 
over  the  method  of  Ex.  1  ? 

114.  Formula.    An  equation  which  expresses  some  prac- 
tical rule  from  arithmetic,  the  shop,  the  trades,  the  sciences, 
the  business  world,  etc.  is  called  a, formula.    Thus,  A—axb 
is  a  practical  formula  for  finding  the  area  of  a  rectangle. 
The  plural  of  "formula"  is  "formulas"  or  "formulae." 

115.  Formula  for  the  area  of  a  square.    The  square  is 
a  special  case  of  a  rectangle ;  that  is,  it  is  a  rectangle  in 
which  a  =  b.    The  formula  can  be  developed  by  the  same 
method   as  for   a  rectangle.    The   only  difference   in  the 
reasoning  is  that  in  every  case  there  are  as  many  rows  of 
square  units  as  there  are  square  units  in  a  row.   (Why  ?) 
Hence  the  formula  for  the  area  of  a  square  is  A  =  b  x  b. 
This  formula  may  be  written  A  =  J2,  where  i2  means  b  x  b, 
and  the  formula  is  read  "A  equals  b  square." 

EXERCISES 

1.  By  the  method  of  counting  squares  find  the  area  of  a 
square  whose  side  is  2|-  cm.  long. 

2.  Apply  the  formula  A  =  Z/2  to  the  square  in  Ex.  1.  .Com- 
pare results. 

3.  Find  the  area  of  a  square  whose  side  is  oft.;  a  feet; 
x  inches  ;  y  yards  ;  m  meters  ;  0.07  mm. ;  2.41  m. 

4.  How  many  feet  of  wire  fencing  are  needed  to  inclose  a 
square  lot  whose  area  is  4900  sq.  ft.  ?    b2  sq.  ft.  ?    4  r2  sq.  yd. '.' 


THE  EQUATION  APPLIED  TO  AREA  79 

5.  Express  by  an  equation  the  area  A  of  a  rectangle  that 
is  8  in.  long  and  5  in.  wide ;   8  in.  long  and  4  in.  wide ;  8  in. 
long  and  3£  in.  wide ;  8  in.  long  and  6^  in.  wide. 

6.  Express  by  an  equation  the  area  A  of  a  rectangle  12  in. 
long  and  of  the  following  widths:  6  in.;  8^  in.;  9^  in.;  10|  in.; 
x  inches ;  y  inches. 

7.  A  mantel  is  54  in.  high  and  48  in.  wide.    The  grate  is 
32  in.  high  and  28  in.  wide.    Find  the  area  of  the  mantel  and 
the  number  of  square  tiles  contained  in  it  if  each  tile  is  3  in. 
on  a  side. 

8.  How  many  tiles  8  in.  square  are  needed  to  make  a  walk 
60  ft.  long  and  4  ft.  wide  ? 

9.  Express  by  equations  the  areas  of  rectangles  1  in.  long 
and  of  the  following  widths  :  12  in. ;  9  in. ;  h  inches  ;  n  inches ; 
x  inches ;  a  inches. 

10.  Express  by  equations  the  areas  of  rectangles  of  width  w 
and  of  the  following  lengths  :   8  ;  10  ;  12^ ;  x ;  a  ;  I ;  b  ;  z. 

11.  In  each  case  write  an  equation  for  the  other  dimension 
of   the   rectangle,   having   given   (a)  altitude  8  in.  and    area 
32  sq.  in. ;  (b)  altitude  5  ft.  and  area  7^  sq.  ft. ;  (c)  base  9  ft. 
and  area  30  sq.  ft. ;  (d)  base  6  in.  and  area  27  sq.  in. ;  (e)  base 
3  in.  and  area  A  square  inches  ;  (f)  base  5  in.  and  area  A  square 
inches  ;  (g)  altitude  a  inches  and  area  A  square  inches  ;  (h)  base 
b  inches  and  area  A  square  inches. 

116.  Formula  for  the  area  of  a  parallelogram.  Fig.  70 
shows  a  parallelogram  that  has  been  transferred  to  that 
position  by  means  of  tracing  paper.  We  wish  to  find  its 
area.  The  line  AB  is  produced  (extended),  and  perpendic- 
ulars are  dropped  from  D  and  C  to  the  line  AB  (see  Art.  84 
for  method  of  constructions),  thus  forming  the  triangles 
AED  inside  and  BFC  adjoining  the  given  parallelogram. 


80 


GEXKi;  A  L    MATHEMATICS 


EXERCISES 

(Exs.  1-7  refer  to  Fig.  70) 

1.  Estimate  by  count  the  number  of  square  units  in  the 
triangle  A  ED. 

2.  Estimate  the  number  of  square  units  in  the  triangle  BFC. 

3.  Compare  the  results  of  Exs.  1  and  2. 

4.  If  the  area  of  the  triangle  .BFC  equals  the  area  of  the 
triangle  A  ED,  what  is  the  relation  between  the  area  of  the 
rectangle  CDEF  and  the  area  of  the  parallelogram  AB CD? 


±- 


FlG.  70.    HOW  TO  FIND  THE  AREA  OF  A  PARALLELOGRAM  BY 
MEANS  OF  SQUARED  PAPER 

5.  What  is  the  formula  for  the  area  of  the  rectangle  CDEF? 
Write  the  formula. 

6.  What  seems  to  be  the  relation  between  the  base  of  the 
parallelogram  and  the  base  of  the  rectangle  ?   What  evidence 
have  you  to  support  your  answer  ? 

7.  What  is  the  relation  between  the  altitude  of  the  paral- 
lelogram and  the  altitude  of  the  rectangle  ?   Give  the  evidence. 

8.  What  seems  to  be  the  formula  expressing  the  area  of 
a  parallelogram  ? 

9.  Without  using  squared  paper  construct  a  parallelogram 
(use  ruler  and  compasses  and  follow  the  method  of  Art.  91). 
Divide  the  parallelogram  into  two  parts — a  triangle  and  a 


THE  EQUATION  APPLIED   TO  AREA  81 

quadrilateral  (as  in  Fig.  71).  Now  shift  the  triangle  to  the 
other  side  so  as  to  form  a  rectangle.  Show  that  the  rectangle 
is  equal  to  the  parallelogram. 

The  preceding  exercises  furnish  evi- 
dence to  support  the  following  law : 
The  area  of  a  parallelogram  equals  the 
product  of  its  base  and  altitude.    This  law  may  be  written 
in  the  form  of  the  following  well-known  formula: 

A  =  a  x  b. 

EXERCISE 

Find  the  area  of  a  parallelogram  if  b '  =  17  in.  and  a  =  5.3  in. ; 
if  b  =  15.4  in.  and  a  —  9.2  in. 

117.  Area  of  a  rhombus.     The  rhombus  (Fig.  72)  is  a 
special  case  of  the  parallelogram,  as  it  is  a  parallelogram 
with  all  its  sides  equal.    Hence  its   area 

equals  its  base  times  its  altitude. 

118.  The  area  of  a  triangle.    The  ex- 
ercises that  follow  will   help   the   pupil 

to   understand  the  formula   for   the   area    of    a  triangle. 

EXERCISES 

(Exs.  1-12  refer  to  Fig.  73) 

1.  Draw  a  triangle  ABC  as  shown  in  Fig.  73. 

2.  Through  C  draw  a  line   CD  parallel  to  AR  (review  the 
method  of  Art.  89). 

3.  Through  B  draw  a  line  parallel  to  AC,  meeting  CD  at  D. 

4.  What  kind  of  a  quadrilateral  is  the  figure  ABDC?  Why  ? 

5.  By  means  of  tracing  paper  transfer  the  parallelogram 
to  squared  paper. 


82  GENERAL  MATHEMATICS 

6.  Estimate  the  number  of  square  units  in  triangle  ABC. 

7.  Estimate  the  number  of  square  units  in  triangle  CBD. 

8.  Compare  the  results  of  Exs.  6  and  7.  What  relation 
does  the  triangle  bear  to  t^e  parallelogram  ? 

9.  What  seems  to  be  the  relation  between  the  base  of  the 
triangle  and  the  base  of  the  parallelogram  ?   Why  ? 

10.  What  is  the  relation  between  the  altitude  of  the  tri- 
angle and  the  altitude  of  the  parallelogram  ?•  Explain'  why. 

11.  What  is  the  formula 
then  for  the  area  of  any 
parallelogram  ? 

12.  What  appears  to  be 
the  formula   for  the   area 

of  a  triangle? 

FIG.  73.    How  TO  FIND  THE  AREA 

13.  Construct     a     paral-  OF  A  TRIANGLE 

lelogram  ABCD.  Construct 

the  diagonal  AC  (a  line  joining  opposite  vertices).  With  a 
sharp  knife  cut  out  the  parallelogram  and  cut  along  the 
diagonal  so  as  to  form  two  triangles.  Try  to  make  one  triangle 
coincide  with  the  other. 

14.  What  conclusion  does  the  evidence  of  Ex.  13  support  ? 

The  preceding  exercises  furnish  evidence  to  show  that 
the  area  of  a  triangle  is  equal  to  one  half  the  product  of 
its  base  and  altitude.  This  law  may  be  written  in  the 
form  of  the  following  formula: 

ab 


I 
119.  Area  of  a  trapezoid.    A  quadrilateral  having  only 

two  sides   parallel  is  called   a  trapezoid  (Fig.  74).    The 
parallel  sides  are  said  to  be  its  bases.    In  Fig.  74  the  upper 


THE  EQUATION  APPLIED  TO  AREA  83 

base  of  the  trapezoid  is  6,  the  lower  base  is  a,  and  the 
altitude  is  h.  To  find  the  area  draw  the  diagonal  BD. 

The  area  of  the  triangle  ABD  =  -.h.    Why ? 

The  area  of  the  triangle  BCD  =  --h.    Why  ? 

2,7 

Therefore  the  area  of  the  trapezoid  =  a '-  +  b  -  •    Why  ? 

77  2t  2 

Note  that  a  —  and  b  -  are  similar  terms.    Why  ?    In  the 

h 
first  term  a  is  the  coefficient  of  —  and  in  the  second  term 

b  is  the  coefficient;  hence,  adding  coefficients,  as  we  may 
always    do    in    adding 
similar  terms,  the  area 

of  trapezoid  is  (a  +  b)  -• 

We  can  only  indicate    A 

the    sum   of    the    two      FlG-  74'  How  J°  FIND  THE  AREA  OF 

A  TRAPEZOID 

bases  until  we  meet  an 

actual  problem.  The  parenthesis  means  that  a  +  b  is  to 
be  thought  of  as  one  number.  The  law  is:  The  area  of 
a  trapezoid  is  equal  to  one  half  the  product  of  its  altitude  by 
the  sum  of  its  bases.  This  law  may  be  written  in  the  form 
of  the  following  formula: 


EXERCISES 

1.  Find  the  area  of  the  trapezoid  whose  altitude  is  12.6  in. 
and  whose  bases  are  8  in.  and  4.6  in.  respectively. 

2.  The  altitude  of  a  parallelogram  is  3  x  +  2,  and  its  base 
is  4  in.    Write  an  algebraic  expression  representing  its  area. 
Find  the  value  of  x  when  the  area  is  28  sq.  in. 


84 


GENERAL  MATHEMATICS 


D 


70 


FIG.  75 


3.  The  altitude  of  a  triangle  is  10  in.,  and  the   hast-  is 
3  x  +  2  in.    Write  an  algebraic  number  representing  the  area. 
Find  the  value  of  ./•  when  the  area  is  55  sq.  in. 

4.  A  man  owns  a  city  lot  with  the  form 
and  dimensions  shown  in  Fig.  75.  ^He  wishes 
to  sell  his  neighbor  a  strip  AEFD  having  a 
frontage  DF  equal  to  10  ft.   If  the  property 
is  worth  $5600,  how  much  should  he  receive 
for  the  strip  '.' 

5.  Of  what  kinds  of  polygons  may  the 
following  equations  express  the  areas  '.' 

(a)  A=x\  (h)  A=l 

(b)  A  =  3«.*-.  _3 

(d)  -i  =5  (a: +  3).  4 

(e)  .4  =  «(«. +  2). 

(f)  .4  =  a(«5  +  4), 

(g)  A  =  *(y  +  2). 

6.  Find  the  value  of  ,1  in  Ex.  5  when  x  =  3,  y  =  2,  a  =  4, 
b  =  1,  and  c  =  5. 

7.  What  quadrilaterals  contain  right  angles  ? 

8.  In  what  respect  does  the  square  differ  from  the  rectangle  ? 

9.  Having  given  a  side,  construct  a  square,  using  only  ruler 
and  compasses. 

HINT.    Review  the  method  for  constructing  a  perpendicular  to  a 
line  segment  (Art.  80). 

10.  Hg>w  does  a  square  differ  from  a  rhombus  ? 

11.  Is  a  rhombus  a  parallelogram  ?   Is  a  parallelogram  a 
rhombus  ? 

12.  Construct  a  rhombus  with  ruler  and  compasses,  given  a 
side  equal  to  5  cm.  and  given  the  included  angle  between  two 
adjacent  sides  as  41°. 

HINT.    Use  the  construction  for  parallel  lines  (Art.  89). 


THE  EQUATION  APPLIED  TO  AREA 


85 


GEOMETRIC  INTERPRETATION  OF  PRODUCTS 

120.  A  monomial  product.    The  formulas  for  the  area  of 
the  rectangle,  the  square,  the  triangle,  the  trapezoid,  etc. 
show  that  the  product  of  numbers 
may  be  represented  geometrically  ; 
for  example,  the  product  of  any 
two  numbers  may  be  represented 
by  a  rectangle  whose  dimensions     FIG.  ~6-  ILLUSTRATING  A 
are  equal  to  the  given  numbers. . 
Thus  the  rectangle  in  Fig.  76  represents  the  product  ab. 


ab 


EXERCISES 

1 .  Sketch  a  rectangle  to  represent  the  product  6  x. 

2.  Sketch  an  area  to  represent  kry. 

3.  Show  from   Fig.  77  that   the   area 
4 x2.    What  is  the  product  of  2 a-  x  2x? 

4.  Show  by  means  of  a  figure  the  area 
of  a  rectangle  3  a  by  5  a. 

5.  Draw  a  figure  to  represent  the  prod-     x 
uct  of  5  a-  and  4  x. 

6.  On  squared  paper  draw  an  area  repre- 

lio.  77.    ILT.USTRAT- 
senting  the  product  ab.   To  the  same  scale    1NG  THE  SQL-ARE  OF 

draw  the  area  ba.    Compare  the  areas.  A  MONOMIAL 

7.  Show  by  a  drawing  on  squared  paper  that  4-5  =  5-4. 

121.  Law  of  order.  The  last  two  exercises  illustrate 
that  in  algebra,  as  in  arithmetic,  the  fen 'torn  of  a  product 
may  be  changed  in  order  without  changing  the  value  of  the 
product.  Thus,  just  as  2x3x5  =  5x3x2,  so  xyz  —  zyx. 
This  is  called  the  Commutative  Law  of  Multiplication. 


i. 

x 

x 

3  expressed 
x         x 

by 

X 

X 

x* 

X              X 

GENERAL  MATHEMATICS 


EXERCISE 

Simplify  the  following :  (a)  2  x  •  3  y  .  4  z ;  (b)  (2  x  y)  (3  x  y} ; 
(c)  4  x  2/  •  3  x  my. 

122.  Product   of   a   polynomial   and   a   monomial.    The 

formula  for  the  trapezoid  suggests  the  possibility  of  draw- 
ing areas  to  represent  the  product  of  a  sum  binomial  by 
a  monomial.  The  process  is  illustrated  by  the  following 
exercises. 

EXERCISES 

1.  Express  by  means  of  an  equation  the  area  of  a  rectangle 
of  dimensions  5  and  x  +  3  (see  Fig.  78).  The  area  of  the 
whole  rectangle  equals  5  (x -\-  3).  D  A  F 

Why?  If  a  perpendicular  be  erected 
at  B  (see  Art.  80  for  method),  the    5 
•rectangle  is  divided   into  two  rec- 
tangles.   The  area  of  DCBA  equals 
5x.     Why?    The    area    of   ABEF 

equals   15.    Why?    It   is    now  easy     PRODUCT  OF  A  POLYNOMIAL 

f.    T    ,,  ,  •  AND  A  MONOMIAL 

to  hnd  the  entire  area  DCEF. 


5x 


15 


C  X  B      3     E 

FIG.  78.   ILLUSTRATING  THE 


Since  DCEF  =  DCBA  +  ABE1*, 

T>  (x  +  3)  =5  x  +  15. 

2.  Show  from  Fig.  79  that  a(x  +  ?/)  =  ax  +  "//. 

3.  Show  by  Fig.  80  that  a  (x  +  T/  +  z)  =  ax  +  a?/  +  az. 


Why  ? 
Why? 


x             y 
FIG.  79 

X                       y                Z 
•FiG.  80 

4.  Draw  an  area  to  represent  bm  +  In  +  be. 

5.  Draw  an  area  to  represent  2  ax  +  2  ay  +  2  az. 

6.  Kepresent  2cc  +  4?/  +  6«byan  area. 

7.  Sketch  a  rectangle  whose  area  equals  2  ax  +  2  ay  +  6  az. 


EUCLID 


88  GENERAL  MATHEMATICS 

• 

HISTORICAL  NOTE.  The  word  "geometry"  comes  from  a  Greek 
phrase  which  means  to  measure  the  earth.  The  early  Egyptians  had 
serious  need  for  a  reliable  method  of  measuring  the  land  after  each 
overflow  of  the  Nile.  The  early  history  of  geometry  appears  to  rest 
on  this  practical  basis. 

The  oldest  collection  of  geometry  problems  is  a  hieratic  papyrus 
written  by  an  Egyptian  priest  named  Ahmes  at  a  date  considerably 
earlier  than  1000  H.C.,  and  this  is  believed  to  be  itself  a  copy  of  some  , 
other  collection  a  thousand  years  older. 

Ahmes  commences  that  part  of  his  papyrus  which  deals  with 
geometry  by  giving  .some  numerical  instances  of  the  contents  of 
barns.  Since  we  do  not  know  the  shape  of  the  barns,  we  cannot 
check  the  accuracy  of  his  work.  However,  he  gave  problems  on 
pyramids.  The  data  and  results  given  agree  closely  with  the  dimen- 
sions of  the  existing  pyramids. 

Geometry  took  definite  form  as  a  science  when  Euclid  (about 
300  B.C.)  wrote  his  "Elements  of  Geometry."  The  proofs  of  his 
text  were  so  excellent  that  the  book  replaced  all  other  texts  of 
the  time  and  has  held  an  influential  position  to  this  day.  The 
form  of  Euclid  is  practically  the  same  as  most  American  geom- 
etry texts,  and  in  England  boys  still  say  they  are  studying  Euclid 
(meaning  geometry). 

We  know  little  of  Euclid's  early  life.  He  may  have  studied  in 
the  schools  founded  by  the  great  philosophers  Plato  and  Aristotle  at 
Athens,  in  Greece.  He  became  head  of  the  mathematics  school 
at  Alexandria,  Egypt,  and  proceeded  to  collect  and  organize  into  a 
set  form  the  known  geometric  principles.  He  is  said  to  have  insisted 
on  the  knowledge  of  geometry  for  its  own  sake.  Thus,  we  read  of 
his  telling  the  youthful  Prince  Ptolemy,  "  There  is  no  royal  road  to 
geometry."  At  another  time,  so  the  story  goes,  when  a  lad  who  had 
just  begun  geometry  asked,  "What  do  I  gain  by  learning  all  this 
stuff  ?  "  Euclid  made  his  slave  give  the  boy  some  coppers,  "  since," 
said  he,  "  he  must  make  a  profit  out  of  what  he  learns." 

Euclid  organized  his  text  so  as  to  form  a  chain  of  reasoning,  begin- 
ning with  obvious  assumptions  and  proceeding  step  by  step  to  results 
of  considerable  difficulty.  The  student  should  read  about  his  work  in 
Ball's  "A  Short  Historyof  Mathematics."  Cajori's"  History  of  Elemen- 
tary Mathematics  "  and  Miller's  "  Historical  Introduction  to  Mathe- 
matical Literature  "  are  further  sources  of  information  about  Euclid. 


THE  EQUATION  APPLIED  TO  AREA  89 

123.  Partial  products.    The  products  ax,  ay,  and  az  in 
the  polynomial   ax  +  ay  4-  az   are   called  partial  products. 
Each  term  of  such  an  expression  may  be  used  to  represent 
the  area  of  some  part  of  a  rectangle.    For  example,  Fig.  80 
shows  a  rectangle  divided  into  the  parts  ax,  ay,  and  az 
respectively.     Here  the  polynomial  ax  +  ay  +  az  may  be 
said  to  represent  the  area  of  the  whole  rectangle. 

124.  Algebraic  multiplication.    The  list  of  exercises  in 
Art.  122  also  shows  that  the  product  of  a  polynomial  and  a 
monomial  is  found  by  multiplying  each  term  of  the  polynomial 
by  the  monomial  and  then  adding  the  partial  products. 

EXERCISES 

1.  Perform  the  following  indicated  multiplications  : 

(a)  3(2  x  +  3  ij).  (c)  3c(2«  +  36). 

(b)  (5x  +  2z)4:a.  (d)3e(2c  +  3). 

2.  Letting  x  =  3,  y  =  1,  and  z  =  2,  find  the  value  of  the 
following  numbers  :  (x  +  y)z;  2  (x  -j-  ?/)  —  2  ;  3  x  +  2  (y  +  z)  ; 

2z);  2s  +  5(x  +  ,/•). 


125.  Geometric  product  of  two  polynomials.  The  product 
of  a  +  c  and  x  +  y  may  be  indicated  as  (a  +  c)  (x  +  y}. 
This  product  may  be  represented 


cy 


,   .     ,,      ,™      <-,.,..    ,  ~  ay 

geometrically  (rig.  ol)  by  a  rec 

tangle   whose    base    is  x  +  y  and 
whose   altitude   is  a  +  c.  FIG.  81.  ILLUSTRATING 

The  rectangle  is  composed  of.       THE  PRODUCT  OF  TWO 
four  rectangles:  ax,  ay,  ex,  and  cy. 

By  the  axiom  that  the  whole  is  equal  to  the  sum  of  its  parts, 
the  whole  rectangle,  («  +  <?)(#+?/),  equals  ax+ay  +  cx  +  cy, 
the  sum  of  the  parts. 


90  GENERAL  MATHEMATICS 

EXERCISES 

1:  Sketch  a  rectangle  whose  area  will  be  the  product  of 
(a  +  ft)  (e  +  d). 

2.  Find  the  geometric  product  of  (c  +  t)  (m  +  n). 

3.  Perform  the  multiplication  (2  -f  x)(m  -f  w)  by  means  of 
a  geometric  figure. 

4.  Find  the  product  (3  x  +  2  ?/)  (a  +  b).    Sketch  the  area 
represented  by  this  product. 

5.  Find  the  product  (a  +  b~)(x  +  ;/  +  z),  using  a  geometric 
figure. 

126.  Algebraic  product  of  two  polynomials.  The  figures 
drawn  in  the  preceding  exercises  indicate  a  short  cut  in 
the  multiplication  of  two  polynomials.  Thus,  a  polynomial 
is  multiplied  by  a  polynomial  by  multiplying  each  term  of 
one  polynomial  by  every  term  of  the  other  and  adding  the 
partial  products. 

EXERCISES 

1.  Using  the  principle  of  Art.  126,  express  the  following 
indicated  products  as  polynomials  : 

(a)  (m  +  n)  (a  +  5).  (g)  3(2  a2  +  a  +  5). 


(f)  5(4  +7  +  3).  (j)  (3x 

(k)  (5  b  +  2  c  +  3  d)  (2  x  +  3  y  +  4  z). 
(1)  (2  m  +  3  n  -f-  ±p]  (3  a  +  7  J  +  5*). 


2.  One  side  of  a  rectangle  is  4yd.  and  the  other  is  6yd. 
How  much  wider  must  it  be  made  so  as  to  be  l£  times  as 
large  as  before?  /• 


THE  EQUATION  APPLIED  TO  AREA  91 

3.  Multiply  the  following  as  indicated  and  check  the  result : 


Solution.    2  x  (x  +  2  y  +  y-)  =  2  x2  +  4  xy  + 
Check.    Let  x  =  2  and  y  —  3. 


2  x2  +  4  xy  +  2  xy"  =  8  +  24  +  36  =  68. 

NOTE.    Avoid  letting  x  =  1,  for  in  this  case  2  x,  2  x2,  2  x8,  etc.  are 
each  equal  to  2.   Why? 

4 .  Multiply  the  following  as  indicated  and  check  the  results  : 
(a) 

(b) 

(c)  (m  +  n  +  a)  (m  +  n  +  &). 

(d)  (m  +  n  +  2 

(e)  (0.4  x  +  0.3  y  +  0.6  s)  (10  x  +  20  y  +  30  z). 

127.  Geometric  square  of  a  binomial.   The  product  of 
(x  +  y*)(x  +  yy,  or  (z  +  y)2,  is  an  interesting  special  case 
of    the    preceding    laws.     The    prod- 
uct may  be  represented  by  a  square 
each    of   whose    sides    is   x  +  y   (see 
Fig.  82).    The  square  is  composed  of 
four  parts,   of  which   two   parts   are 
equal.   Since  these  two  parts  are  repre- 
sented by  similar  algebraic  terms,  they 
may  be  added ;  thus,  xy  -\-xy-2  xy.  x  V 

Hence   the   area  of   a  square  whose     FIG.  82.  ILLUSTRATING 
•  j       •  o   .    r>  o      rru         THE   SQUARE  OF  A  Bi- 

side  isaj-t-yisar+Jsa^  +  ^r.     1  he 

same  product  is  obtained  by  apply- 
ing the  law  for  the  product  of  two  polynomials ;  thus, 

x  +y 

x  +y 


xy 


xy 


+ 


xy 

xy  + 


2  xy 


92  GENERAL  MATHEMATICS 

In  algebraic  terms  we  may  say  that  the  square  of  the 
sum  of  tivo  numbers  equals  the  square  of  the  first,  plus  twice 
the  product  of  the  two  numbers,  plus  the  square  of  the  second. 

Use  Fig.  82  to  show  what  this  law  means. 

EXERCISES 

1.  By  means  of  figures  express  the  following  squares  as 
polynomials  : 


(b)  (m  +  n)2.  (e)  (x  +  2)2.  (h)  (2  x  +  y)2. 

(c)  (c  +  d)*.  (f  )  (m  +  3f.  (i)  (2  x  +  3  y)2. 

2.  Sketch   squares    that   are   suggested    by  the    following 
trinomials  : 

(a)  a2  +  2  ah  +  lr.  (e)  mz  +  8  MI  +  16. 

(b)  x2  +  2  ax  +  a2.  (f)  .r2  +  10  a-  +  25. 

(c)  A-2  +  2  A->-  -(-  r.  (g)  49  +  14  x  +  z2. 

(d)  x2  +  4*  +  4.  (h)  c2  +  c  +  ], 

3.  Indicate  what  number  lias  been  multiplied  by  itself  to 
produce 

(a)  or2  +  2  vy  +  /.  (c)  z2  +  6  a-  +  9. 

(b)  r2  +  4  r  +  4.  (d)  &a  +  10  //  +  25. 

4.  What  are  the  factors  in  the  trinomials  of  Ex.  .3  '.' 

5.  The  'following  list  of  equations  review  the  fundamental 
axioms  as  taught  in  Chapter  I.   Solve  each  equation  and  check 
by  the  methods  of  Chapter  I. 

(a)  30  +  4)  =22  +  ,-. 

(b)  9  0  +  35)  =5  (2  a  +  45). 

(c)  3(x  +  15)+  5  =  2(2*  +  9)  +  4(.r  +  3). 

(d)?fe±2-8.         W|i-|  =  1.  (h)f-f=a 

2.   fc,  5  +  5-2.    (i)^^  =  8. 


. 


93 

128.  Evaluation.    The  area  of  each  of  the  geometrical 
figures  considered  in  this  chapter  has  been  found  to  depend 
upon  the  dimensions  of  the  figure.    This  dependence  has 
been  expressed  by  means  of  formulas,  as  A  =  ab  in  the  case 
of  the  rectangle.    Whenever  definite  numbers  are  substi- 
tuted in  the  expression  ab  in  order  to  find  the  area,  A, 
for  a  particular  rectangle,  the  expression  ab  is  said  to  be 
evaluated.    This  process   implies  getting  practical  control 
of  the  formulas. 

EXERCISES 

1.  Find    the    value    of   A    in    the    formula    .1  =  ab    when 
a  =  22.41  ft.  and  b  =  23.42  ft. 

2.  Find    the    value    of    J     in    the    formula   A  =—    when 
a  =  12.41  ft.  and  I  =  2.144  ft. 

3.  Find  the  value  of  .1  in  the  formula  A  =  (<i  -)-  fi)  —  when 
a  =  12.42  ft,  b  =  6.43  ft.,  and  h  =  20.12  ft. 

129.  The  accuracy  of  the  result.    In  finding  the  area  in 
Ex.  1  above  we  get  A  =  (22.41)  (23.42)=  524.8422  sq.  ft. 
This  is  a  number  with  four  decimal  places.    As  it  stands 
it  claims  accuracy  to  the  ten-thousandth  of  a  square  foot. 
The  question  arises  whether  this  result  tells  the  truth. 

Suppose  the  numbers  above  represent  the  length  and  width  respec- 
tively of  your  classroom.  Does  the  product  524.8422  sq.  ft.  indicate 
that  we  actually  know  the  area  of  the  floor  accurate  to  one  ten- 
thousandth  of  a  square  foot?  Shall  we  discard  some  of  the  deci- 
mal places  ?  If  so,  how  many  are  meaningless  ?  How  much  of  the 
multiplication  was  a  waste  of  time  and  energy  ?  These  questions  are 
all  involved  in  the  fundamental  question  How  many  decimal  places 
shall  we  regard  as  significant  in  the  process  of  multiplication  ?  . 

It  is  important  that  we  have  a  clear  understanding  of 
the  question.  For  if  we  carry  along  in  the  process  mean- 
ingless decimals  we  are  wasting  time  and  energy,  and, 


94 

what  is  more  serious,  we  are  dishonestly  claiming  for 
the  result  an  accuracy  which  it  does  not  have.  On  the 
other  hand,  we  are  not  doing  scientific  work  when  we 
carelessly  reject  figures  that  convey  information. 

The  following  facts  are  among  those  which  bear  on  our 
problem : 

(a)  In  Art.  26  we  pointed  out  that  any  number  obtained 
by  measurement  is  an  approximation.'   The  application  of 
the  area  formulas  involves  the  measurement  of  line  seg- 
ments.   Hence   an  area  is   an   approximation.    This   fact 
alone  is  sufficient  to  make  us  exceedingly  critical  of  the 
result  524.8422  sq.  ft.  as  an  absolutely  accurate  measure 
of  the  area  of  the  classroom  floor. 

(b)  If  we  measure  the  length  of  a  room  with  a  reliable 
tape  measure  and  record  the  result  as  23.42  ft.,  this  does 
not  mean  that  we  regard  the  result  as  absolutely  accurate. 
If  the  scale  is  graduated  to  hundredths  of  a  foot,  it  means 
that  23.42  ft.  is  the  result  nearest  to  the  true  value.    The 
eye  tells  us  that  23.425  ft.  is  too  high  and  23.415  ft.  is 
too  low,  but  that  the  result  may  be  anywhere  between 
these.    Thus,  the  length  of  the  room  lies  anywhere  between 
23.415  ft.   and    23.425  ft.    Similarly,   the  width  may  be 
anywhere  between  22.405  ft.  and  22.415  ft.    The  student 
should  practice  measuring  objects  with  a  yardstick  or  a 
meter  stick  till  the  point  of  this  paragraph  is  clear  to  him. 
Test  question :    How  does  2.4  ft.  differ  from  2.40  ft.  ? 

Multiplying  the  smallest  possible  length  (23.415  ft.) 
of  the  classroom  by  the  smallest  width  (22.405  ft.)  we 
get  a  possible  area  of  524.613075  sq.  ft.  By  multiplying 
the  greatest  length  (23.425  ft.)  by  the  greatest  width 
(22.415  ft.)  we  get  525.071375  sq.  ft.  Subtracting  the 
smallest  possible  area  from  the  largest  possible  area  gives 
us  a  range  of  over  0.45  of  a  square  foot.  In  short,  the 


THE  EQUATION  APPLIED  TO  AREA  95 


result  might  be  wrong  by  practically  one  half  of  a  square 
foot.  We  are  not  actually  sure  of  the  third  figure  from  the 
left.  It  mayfce  a  4  or  a  5.  We  shall  be  reasonably  near 
the  truth  if  w%  record  the  result  simply  as  524.?  sq.  ft., 
a  number  chosen  roughly  halfway  between  the  largest  and 
smallest  possible  areas. 

It  can  thus  be  shown  that  the  product  of  two  approxi- 
mate four-place  numbers  is  not  to  be  regarded  accurate 
to  more  than  four  places. 

*130.  Abbreviated  multiplication.  It  is  apparent  in  the 
preceding  discussion  that  it  is  a  waste  of  time  to  work 
out  all  the  partial  products  in  multiplication.  It  is  easier 
(when  the  habit  is  once  established)  to  work  out  only 
the  partial  products  which  go  to  make  up  the  significant 
part  of  the  answer. 

Thus,  47.56  x  34.23  may  take  the  following  forms : 

ABBREVIATED  FORM  USUAL  METHOD 

47.56  By  multiplication  we  get 

34.23  47.56 

1427  34.23 

190 
10 
1 
1628. 


The  difference  is  accidentally  only  a  little  more  than 
0.02  sq.  ft.  It  can  be  shown  by  the  method  used  in  the 
classroom  problem  (Art.  129)  that  1628  is  easily  in  the 
range  of  probable  areas ;  that  is,  we  are  not  actually  sure 
about  the  fourth  figure  from  the  left. 

*  Hereafter  all  articles  and  exercises  marked  with  an  asterisk  may  be 
omitted  without  destroying  the  sequence  of  the  work. 


1 

4268 

9 

512 

190 

24 

1426 

8 

1627 

9788 

96  GENERAL  MATHEMATICS 

The  abbreviated  method  consists  of  writing  only  the 
significant  parts  of  the  usual  method  (see  numbers  to  left 
of  the  line).  Add  1  unit  when  the  number^o  the  right  is 
the  figure  5  or  larger.  The  method  will  pppear  awkward 
until  sufficiently  practiced. 

A  similar  discussion  concerning  accuracy  could  be  given 
for  division.  In  addition  or  subtraction  it  is  easy  to 
see  that  the  sum  or  difference  of  two  numbers  cannot  be 
regarded  as  more  accurate  than  the  less  accurate  of  the 
two  numbers.  Illustrate  the  truth  of  the  last  statement. 

While  the  discussion  of  this  very  important  topic  has 
been  by  no  means  complete,  perhaps  enough  has  been 
said  to  fulfill  our  purpose,  which  is  to  make  the  student 
exceedingly  critical  of  results  involving  the  significance  of 
decimal  places. 

EXERCISES 

*1.  Assuming  that  the  dimensions  of  a  hall  are  measured 
with  a  reliable  steel  tape  and  that  the  dimensions  are  recorded 
as  47.56  ft.  and  34.23  ft.  respectively,  show  by  the  method  used 
in  Art.  129  that  the  difference  between  the  smallest  and  the 
largest  possible  area  of  the  hall  is  actually  over  four  fifths  of 
a  square  foot. 

*2.  By  means  of  the  abbreviated  multiplication  method 
write  the  product  of  46.54  and  32.78 ;  of  23.465  and  34.273. 

*3.  Multiply  by  the  usual  method  and  compare  the  short- 
cut result  with  this  result. 

*4.  Which  result  is  the  more  accurate  ? 

SUMMARY 

131.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  area,  measuring  area,  unit  of  area, 
rhombus,  trapezoid.  Commutative  Law  of  Multiplication, 
partial  products,  parenthesis,  formula,  formulas. 


THE  EQUATION  APPLIED  TO  AREA  97 

132.  The  following  formulas  have  been  taught: 

(a)  A  =  ba.    (For  the  area  of  a  rectangle.) 

(b)  A  =  b2.    (For  the  area  of  a  square.) 

(c)  A=bh.    (For  the  area  of  a  parallelogram.) 

(d)  A  =  —  •    (For  the  area  of  a  triangle.) 

7 

(e)  A  =  (a  +  £>)  -  •    (For  the  area  of  a  trapezoid.) 

a 

(f  )  (x  +  #)2=  y?+  2  xy  +  3/2.    (For  the  area  of  a  square 
whose  side  is  x 


133.  The  product  of  two  numbers  may  be  represented 
geometrically  as  an  area. 

134.  The  algebraic  product  of  a  monomial  and  a  poly- 
nomial is  found  by  multiplying  each  term  of  the  polynomial 
by  the  monomial  and  then  adding  the  partial  products. 

135.  The  product  of  two  polynomials  is  the  sum  of  all 
the  partial  products  obtained  by  multiplying  each  term 
of  one  polynomial  by  every  term  of  the  other. 

*136.  We  need  to  be  very  critical  of  the  number  of 
decimal  places  that  we  submit  in  a  result.  The  product  of 
two  approximate  four-digit  numbers  is  only  approximately 
correct  for  four  digits. 


CHAPTER  V 

THE  EQUATION  APPLIED  TO  VOLUME 

137.  Solids.  The  drawings  in  Fig.  83  represent  geo- 
metric solids.  A  solid  is  commonly  thought  of  as  an  object 
that  occupies  a  portion  of  space.  It  is  separated  from 


Oblique  Paral- 
lelepiped 


Cube 


Rectangular  Paral- 
lelepiped 


Triangular 
Pyramid 


Sphere 
FIG.  83.    FAMILIAR  SOLIDS 


Frustum  of  a 
Pyramid 


the  surrounding  space  by  its  surface.  In  geometry  we 
study  only  the  form  of  the  solid  and  its  size.  We  are 
not  interested  in  color,  weight,  etc.  A  solid  differs  from 
the  figures  we  have  been  studying  in  that  it  does  not  lie 
altogether  in  a  plane,  but  involves  a  third  dimension.  What 
figures  in  two  dimensions  are  suggested  by  the  solids  in 
Fig.  83  ?  For  example,  the  square  is  suggested  by  the  cube. 

98 


THE  EQUATION  APPLIED  TO  VOLUME 


99 


138.  Cube.  The  cube    has  six    faces  all   of  which  are 
squares.  Two  faces  intersect  in  an  edge.    How  many  edges 
has  a  cube  ?  How  many  corners  ?  How  is  a  corner  formed  ? 

139.  Oblique   parallelepiped.    The  faces  of    an    oblique 
parallelepiped    are    all   parallelograms.    How  many  faces 
has  it?    How  many  vertices?    How  many  edges? 

140.  Rectangular  parallelepiped.     The   faces   of   a  rec- 
tangular parallelepiped  are  rectangles. 

141.  Measurement  of  volume  ;  unit  of  volume.  When  we 
determine  the  amount  of  space  inclosed  within  the  surface 
of  a  solid  we  are  measuring  the  volume  of  the  solid.    To 
measure  the  volume  of  a  solid  we  compare  the  solid  with 
a  cube  each  of  whose  edges  equals  a  unit  of  length.    The 
volume  is  expressed  numerically  by  the  number  of  times 
the  unit  cube  goes  into  the  solid.    The  unit  cube  is  called 
the  unit  of  volume. 

142.  Formula  for  the  volume  of  a  rectangular  parallele- 
piped.   In  Fig.  84  a  rectangular  parallelepiped  is  shown 
which  is  5  cm.  long,  3  cm.  wide, 

and  4  cm.  high.    The  unit  cube 

is  represented  by  K.    Since  the 

base    of    the    solid    (the    face 

on  which  it  stands)  is   5  cm. 

long  and  3  cm.  wide,  a  layer  of 

3  x  5  unit  cubes  could  be  placed 

upon  it.   Since  the  solid  is  4  cm. 

high,  it  contains  4  layers  of  unit 

cubes  ;  that  is,  4x3x5,  or  60, 

unit  cubes.    Thus  the  volume 

of  a  rectangular  parallelepiped 

is  obtained  by  multiplying  the  length  by  the  width  by  the 

height.    This  law  may  be  expressed  by  the  formula  V=  Iwh. 


B 


/     / 


7 


N 


R 

FIG.  84.    How  TO  FIND  THE 
VOLUME  OF  A  RECTANGULAR 
PARALLELEPIPED 


100 


EXERCISES 

1.  Find  the  volume  of  a  rectangular  parallelepiped  if  its 
dimensions  are  I  =  63  in.,  h  =  42  in.,  and  w  —  56  in. 

*2.  If  in  the  preceding  discussion  the  edges  of  the  rec- 
tangular parallelepiped  had  not  been  given  as  integral  multiples 
of  the  unit  cube,  it  would  have  been  necessary  temporarily  to 
adopt  a  smaller  unit  cube.  Show  that  the  formula  V  =  Iwh 
holds  when  I  =  2.3  cm.,  h  —  3.4  cm.,  and  w  =  1.7  cm. 
HINT.  Follow  the  method  suggested  in  Art.  111. 

*3.  Show  by  means  of  a  general  discussion  that  the  formula 
would  be  true  if  I  =  3j,  h  =  2j,  w  =  3f . 
HINT.    See  Ex.  3,  Art.  112. 

*  143.  Volume  of  an  oblique  parallelepiped.  Fig.  85  shows 
in  a  general  way  the  method  used  in  a  more  advanced 
mathematics  course  to  show  that  the  formula  V=  Iwh  holds 


in 


}h 


R 


FIG.  85.    MODEL    ILLUSTRATING    HOW    TO   FIND   THE   VOLUME    OF   AN 
OBLIQUE  PARALLELEPIPED 

even  for  aji  oblique  parallelepiped.  Parallelepiped  III  is  a 
rectangular  parallelepiped,  and  we  know  the  formula  holds 
for  it.  Parallelepiped  II  is  a  right  parallelepiped  (it  has 


THE  EQUATION  APPLIED  TO  VOLUME       101 

four  rectangular  faces,  and  two  are  parallelograms)  and 
by  advanced  methods  is  shown  equal  to  parallelepiped  III. 
Parallelepiped  I  is  oblique  and  is  shown  equal  to  parallele- 
piped II.  Since  parallelepiped  I  equals  parallelepiped  II, 
and  parallelepiped  II  in  turn  equals  parallelepiped  III,  the 
formula  holds  for  parallelepiped  I.  The  student  should  not 
be  concerned  if  he  cannot  fully  understand  this  discussion. 
He  should  be  ready  to  apply  the  formula  for  an  oblique 
parallelepiped  when  the  need  for  it  arises  in  shop  or  factory 
just  as  he  does  many  principles  of  arithmetic. 

EXERCISES 

*1.  It  will  be  easy  for  some  student  to  make  models  of  the 
preceding  figures  in  the  shop.  Thus,  to  show  parallelepiped  II 
equal  to  parallelepiped  III  construct  parallelepiped  II  and 
drop  a  perpendicular  from  D  to  the  base.  Then  saw  along  the 
edges  MD  and  DF.  Place  the  slab  obtained  on  the  right  side, 
and  parallelepiped  II  will  look  exactly  like  parallelepiped  III. 
This  will  be  helpful  to  your  classmates,  and  you  will  find  the 
exercise  easy  and  interesting. 

*2.  A  much  more  difficult  and  interesting  exercise  is  to  make 
parallelepiped  I  look  like  parallelepiped  III. 

HINT.  Construct  RK±  to  AC  and  KI±  to  AC.  Saw  along  the 
edges  RK  and  KI  and  place  the  slab  obtained  on  the  left  side.  Now 
the  figure  will  be  transformed  to  parallelepiped  II.  Continue  as  in 
Ex.  1  to  make  parallelepiped  I  look  like  parallelepiped  III. 

3.  A  rectangular  reservoir  120  ft.  long  and  20  ft.  wide  con- 
tarns  water  to  a  depth  of  10.5  ft.  A  second  reservoir  125  ft.  long 
and  18  ft.  wide  contains  water  to  a  depth  of  36.5  ft.   How  much 
more  water  is  there  in  the.  second  reservoir  than  in  the  first  ? 

4.  A  rectangular  tank  6  ft.  long,  4  ft.  wide,  and  5  ft.  deep 
is  to  be  lined  with  zinc  ^  in.  thick.    How  many  cubic  feet  of 
zinc  will  be  required  if  4  sq.  ft.  are  allowed  for  overlapping  ? 


102  GENERAL  MATHEMATICS 

5.  If  1  cu.  in.  of  pure  gold  beaten  into  gold  leaf  will  cover 
30,000  sq.  ft.  of  surface,  what  is  the  thickness  of  the  gold  leaf  ? 

6.  An  open  tank  is  made  of  iron  ^  in.  thick.    The  outer 
dimensions  are  as   follows  :    length,  3  ft. ;  width,  1  ft.  9  in. ; 
height,  2  ft.    If  1  cu.  ft.  of  iron  weighs  460  lb.,  find  the  weight 
of  the  tank. 

7.  In  a  rainfall  of  1  in.  how  many  tons  of  water  fall  upon 
an  acre  of  ground  if  1  cu.  ft.  of  water  weighs  62.5  lb.  ? 

144.  Formula  for  the  volume  of  a  cube.    The  volume  of 
a  cube  is  computed  in  the  same  way  as  that  of  a  parallele- 
piped.   The  cube  is  a  special  case  in  the  sense  that  the 
length,  width,  and  height  are  all  equal.  Hence,  if  s  equals  an 
edge  of  a  cube,  the  volume  may  be  expressed  by  the  formula 
V  =  s  x  s  x  s.    The  formula  V=  s  x  s  x  s  may  be  written 
more  briefly  V=  s3  (read  "V  equals  s  cube");  s3  being  an 
abbreviated  form  of  s  x  s  x  s.    The  formula  may  be  trans- 
lated into  the  following  law :    The  volume  of  a  cube  equals 
the  mbe  of  an  edge. 

EXERCISES 

1.  Find  the  volume  of  a  cube  whose  edge  is  ^  in. ;  ^  in. ;  ^  in. 

2.  Find  the  volume  of  a  cube  whose  edge  is  1^  in.;  2.2  cm.; 
3  in.;  1  m. ;  0.01  m. 

145.  Equal  factors  ;  exponents  ;  base  ;  power.   The  prod- 
ucts   of   two    equal    dimensions   and    three   equal   dimen- 
sions have  been  represented  by  the  area  of  a  square  and 
the  volume  of  a  cube  respectively.    Hence  the  notation 
"  a  square  "  and  "s  cube."    The  product  of  four  equal  fac- 
tors cannot  be  represented  geometrically,  though  you  may 
already  have  heard  people  talk  vaguely  about  the  fourth 
dimension.    However,  the  product  of  four  equal  algebraic 
factors,  say  s  x  s  x  s  x  s,   is   as  definite  as   2x2x2x2 


.  THE  EQUATION  APPLIED  TO  VOLUME       103 

in  arithmetic.  Thus,  we  extend  the  process  indefinitely  in 
algebra  and  write  sxsxsxs  =  s*  (read  "s  fourth")  or 
bxbxbxbxb  =  b*  (read  "  b  fifth  "),  etc. 

The  term  b5  is  obviously  much  more  convenient  than 
bxbxbxbxb.  The  5  hi  b6  is  called  an  exponent.  It 
is  a  small  number  written  to  the  right  and  a  little  above 
another  number  to  show  how  many  times  that  number  is  to 
be  used  as  a  factor.  In  53  (meaning  5x5x5)  the  3  is 
the  exponent,  the  number  5  is  the  base,  and  the  product 
of  5  x  5  x  5  is  the  power.  Thus,  125,  or  53,  is  the  third 
power  of  5.  When  no  exponent  is  written,  as  in  x,  the 
exponent  is  understood  to  be  1.  Thus,  in  2  xy,  both  x  and 
y  are  each  to  be  used  only  once  ast  a  factor.  The  mean- 
ing is  the  same  as  if  the  term  were  Mjritten  2xlyl. 

EXERCISES 

1.  State  clearly  the  difference  between  a  coefficient  and  an 
exponent.   Illustrate  with  arithmetical  numbers. 

2.  Letting  a  =  5,  give  the  meaning  and  the  value  of  each 
of  the  following  numbers : 

(a)  3  a.      (c)  2  a.       (e)  4  a.       (g)  5  a.       (i)  2  a2.        (k)  4  a2. 

(b)  a8.        (d)  a?.        (f)  a4.         (h)  a5.          (j)  3  a2.        (1)  2  a4. 

3.  Write  the  following  products  in  briefest  form : 

1111        333  a    a    a 

yyyyy,    5-5-5-5;    g-^j—i     m'> 

2J  .  5J  •  2J ;  5 . 4  V4f  •  6  •  3  ^  3  •  3  •  *  • « .  y .  y .  y. 

4.  Find  the  value  of  2s;  68;  (i)4;  38;  (1.3)a;  98;  (0.03)8; 
(1.1)8. 

5.  Letting  m  =  2  and  n  =  3,  find  the  value  of  the  follow- 
ing polynomials  :    m2  +  2  mn  +  w2 ;    m"  +  3  ra2n,  +  3  wp2  +  w8 ; 

);   5(m  +  n);   6(2  ws  +  3m2  -f-  4mn  +  n  +  3). 


104  GENERAL  MATHEMATICS 

6.  Find  the  value  of  the  following  numbers,  where  z  —  3 : 
(a)  2z;    (b)  *»;    (c)  (2*)2;    (d)  2*2;    (e)  3z8;    (f)  (3z)3; 

(g)  (3*)2- 

7.  Letting  a;  =  1,  y  =  2,  z  =  3,  and  u  =  4,  find  the  value  of 
the  following : 

xy  +  xz  +  yu  +  zu    x*  +  4  x3#  -f-  6  x2y2  4-  4  x^3  +  2/4 
#  -f  y  +  «  +  w  xy 

146.  Exponents  important.    Since  the  subject  of  expo- 
nents is  fundamental  to  a  clear  understanding  of  two  very 
important  labor-saving  devices,  namely  the  slide  rule  and 
logarithms,  which  we  shall  presently  study,  it  is  necessary 
to  study  the  laws  of  exponents  very  carefully. 

147.  Product  of  powers  having  the  same  exponents.    The 
law  to  be  used  in  this  type  may  be  illustrated  by  the 
problem,  "  Multiply  a2  by  a3."    The  expression  a2  means 
a  •  a,  or  aa,  and  the  expression  a?  means  a  •  a  •  a,  or  aaa. 
Hence  a2  •  a3  means  aa  •  aaa,  or,  hi  short,  a6. 

EXERCISE 

In  each  case  give  orally  the  product  in  briefest  form  : 

(a)  32-33.  (i)  xs-  x6.  (q)  x2  •  m. 

(b)  6-68.  (j)  ax-x.  (r)6-6.*>fl£ 

(c)  52  •  5*.  (k)  b  •  e  •  b.  (  s  )  4  tfc  •  5  iV. 

(d)  10  •  108.  (1)  b-b.  ( ;  t )  xy»-  x2yz2  •  2  xifzs. 

(e)  x  •  x2.  (m)   e  •  e  •  2  •  2.  (u)  (2  xyf. 

(f)  122.12S.          (n)  c-c3.  (v)   (2xV)3. 

(g)  x2  •  xs.  (  o  )  x  •  x5.  (w) '  (3  x2?/)2. 

(h)  x  •  x4.  (p)  m  •  a-2.  (x)  3  •  5  •  2  •  5  •  2  •  38. 

The  exercise  above  shows  that  the  product  of  two  or 
more  factors  having  the  same  base  is  a  number  ivJwse  base 
is  the  same  as  that  of  the  factors  and  whose  exponent  is  the 
mm  of  the,  exponent*  of  the  factors ;  thus,  b2  •  &  .  /£  =  2>10. 


THE  EQUATION  APPLIED  TO  VOLUME       105 

148.  Quotient  of  powers  having  the  same  base.  The  quo- 
tient of  two  factors  having  the  same  base  may  be  simplified 
by  the  method  used  in  the  following  problem: 

Divide  b6  by  b\ 

HINT.   Since  a  fraction  indicates  a  division,  this  quotient  may 

65 

be  indicated  in  the  form  —  • 
cr 

b5     b-b'b-b'b 

Solution.  -  =  -  —  -  -- 

&•*  b    b 

Hence,  dividing  numerator  and  denominator  by  b  •  b,  or  b2, 


EXERCISE 

In  each  case  give  orally  the  quotient  in  briefest  form  : 

2»  (c)ft'-*-P.  (g)*6^*2. 

w  2"  '  (d)  b6  --  b\  (h)  x6  -4-  xs. 

5_«  (e)b"  +  b*.  (i)x°  +  x*. 

W  52'  (f)  b™  +  bv  (j)  x21  +  x". 

The  preceding  exercise  shows  that  the  quotient  obtained 
by  dividing  a  power  by  another  power  having  the  same  base 
is  a  number  whose  base  is  the  common  base  of  the  given 
powers  and  whose  exponent  is  obtained  by  subtracting  the 
exponent  of  the  divisor  from  the  exponent  of  the  dividend. 

Thus,  a:10  •*•  x3  =  x1.  Here  x1  has  the  same  base  as  the  dividend  xw 
and  the  divisor  Xs  ;  and  its  exponent,  7,  is  obtained  by  subtracting  3 
from  10. 

149.  Review  list  of  problems  involving  application  of 
algebraic  principles  to  geometric  figures.  The  following 
exercises  are  intended  to  help  the  student  to  see  how  to 
apply  algebraic  principles  to  geometric  figures. 


106 


2X+3 


FIG. 


EXERCISES 

1.  Find  an  algebraic  number  which  will  express  the  sum 
of  the  edges  of  the  solid  in  Fig.  86. 

2.  If  the  sum  of  the  edges  of  the  solid  in  Fig.  86  is  172,  what 
are  the  actual  dimensions  of  the  solid  ? 

3.  Find     an     algebraic     expression 
for   the    total    surface    of   the   solid    in 
Fig.  86.    Also  for  the  volume. 

4.  What  is  the  total  surface  and  vol- 
ume of  the  solid  in  Fig.  86  if  x  equals  10  ? 

5.  Express  algebraically  the  sum  of  the  edges  of  the  cube 
in  Fig.  87. 

6.  If  the  sum  of  the  edges  of  the  culx- 
in    Fig.  87    is    112,    what    is    the    length   of 
one  edge  ? 

7.  Express  algebraically  the  total  surface 
and  volume  of  the  cube  in  Fig.  87. 

8.  What  is  the  total  surface  of  the  cube  in  Fig.  87  if  x  =  2  ? 

9.  The   edge   of   a  tetrahedron  (Fig.  88)  is   denoted   by 
2  x  + 1.    Express    algebraically  the    sum  of  all 

the  edges  of  the  tetrahedron. 

NOTE.  A  tetrahedron  is  a  figure  all  of  whose 
edges  are  equal  and  whose  faces  are  equal  equilateral 
triangles. 

10.  Find  the  length  of  an  edge  of  the  tetra- 
hedron   in    Fig.  88    if   the   sum    of   the    edges    is   40.5  cm. 

11.  Fig.  89  shows  a  frustum  of  a  pyra- 
mid.  The  upper  and  lower  bases  are  equi- 
lateral pentagons  ;  the  sides  are  trapezoids 
with  the  edges  denoted  as  in  the  figure. 
Find   the    sum    of   all    the    edges.    If   e 

equals  2.  what  is  the  sum  of  the  edges  ?  FIG.  89 


DESCARTES 


108  GENERAL  MATHEMATICS 

HISTORICAL  NOTE.  The  idea  of  using  exponents  to  mark  the 
power  to  which  a  quantity  was  raised  was  due  to  Rene1  Descartes, 
the  French  philosopher  (1596-1650).  It  is  interesting  to  read  of  the 
struggle  for  centuries  on  the  part  of  mathematicians  to  obtain 
some  simple  method  of  writing  a  power  of  a  number.  Thus,  we  read 
of  the  Hindu  mathematician  Bhaskara  (1114-  )*using  the  initials 
of  the  Hindu  words  "  square  "  and  "  solid  "  as  denoting  the  second  and 
third  power  of  the  unknown  numbers  in  problems,  which  he  gave  a 
practical  setting  with  many  references  to  fair  damsels  and  gallant 
warriors.  In  the  following  centuries  a  great  variety  of  symbols  for 
powers  are  used ;  for  example,  arcs,  circles,  etc.,  until  we  come  to  a 
French  lawyer,  Frangois  Vieta  (1540-1603),  who  wrote  on  mathe- 
matics as  a  pastime.  Vieta  did  much  to  standardize  the  notation  of 
algebra.  Thus,  in  the  matter  of  exponents  he  employed  "  A  quad- 
ratusx"  "  A  cubus,"  to  represent  z2  and  x8,  instead  of  introducing  a 
new  letter  for  each  power.  From  this  point  it  is  only  a  step  to 
Descartes's  method. 

The  biographies  of  the  three  mathematicians  Bhaskara,  Viet;a,  and 
Descartes  are  exceedingly  interesting.  Thus,  you  may  enjoy  reading 
of  Bhaskara's  syncopated  algebra  in  verse,  in  which  many  of  the 
problems  are  addressed  to  "  lovely  and  dear  Lilavati "  (his  daughter) 
by  way  of  consolation  when  he  forbade  her  marriage. 

You  may  read  of  Vieta's  being  summoned  to  the  court  of  Henry  IV 
of  France  to  solve  a  problem  which  involved  the  45th  power  of  x. 
The  problem  had  been  sent  as  a  challenge  to  all  mathematicians 
in  the  empire.  Vieta  appeared  in  a  few  moments  and  gave  the  king 
two  correct  solutions.  Next  King  Henry  asked  Vieta  to  decipher 
the  Spanish  military  code,  containing  over  six  hundred  unknown 
characters,  which  was  periodically  changed.  King  Henry  gave  the 
cipher  to  Vieta,  who  succeeded  in  finding  the  solution  to  the  system, 
which  the  French  held  greatly  to  their  profit  during  the  war. 

Or  you  may  read  of  Descartes,  a  member  of  the  nobility,  who 
found  the  years  of  his  army  life  exceedingly  irksome,  for  he  craved 
leisure  for  mathematical  studies.  He  resigned  his  commission  in 
1621  and  gave  his  time  to  travel  and  study.  In  1637  he  wrote  a  book, 
"  Discourse  on  Methods."  In  this  text  he  made  considerable  advance 
toward  the  system  of  exponents  now  used.  The  text  shows  that  he 
realized  the  close  relation  existing  between  geometry  and  algebra. 
He  is  often  called  "the  father  of  modern  algebra." 


THE  EQUATION  APPLIED  TO  VOLUME       109 


12.  A  tetrahedron  may  be  constructed  from  a  figure  like 
Fig.  90.  Draw  the  figure  on  cardboard,  using  a  larger  scale. 
Cut  out  the  figure  along  the  heavy  lines ;  then  fold  along 
the  dotted  lines.  Join  the  edges  by  means  of  gummed  paper. 


FIG.  90.    How  TO  CON- 
STRUCT A  TETRAHEDRON 


FIG.  91.   How  TO  CONSTRUCT 
A  CUBE 


13.  The  cube  may  be  constructed  from  a  figure  like  Fig.  91. 
Draw  the  figure  on  cardboard,  using  a  larger  scale ;  for  example, 
let  x  =  3  cm.   Cut  out  the  figure 

along  the  heavy  lines,  then  fold 
along  the  dotted  lines.  Join  the 
edges  by  means  of  gummed  paper. 
This  will  form  a  model  of  a  cube. 

14.  Measure  the  edge  of  the 
cube  constructed  for  Ex.  13  and 
compute  the  area  of  the  whole 
surface.    Find  the  volume  also. 

15.  A    rectangular    parallele- 
piped may  be  constructed  from 

a  figure  like  Fig.  92.   Compute  the  volume  of  the  solid  and  the 
area  of  the  surface. 

SUMMARY 

150.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  a  solid,  surface  of  a  solid,  volume 
of  a  solid,  unit  of  volume,  cube,  parallelepiped,  rectangular 
parallelepiped,  right  parallelepiped,  oblique  parallelepiped, 
triangular  pyramid,  exponent,  base,  power,  tetrahedron. 


FIG.  92.   How  TO  CONSTRUCT  A 
RECTANGULAR  PARALLELEPIPED 


110  GENERAL  MATHEMATICS 

151.  The  volume  of  a  solid  is  determined  by  applying 
the  unit  cube  to  see  how  many  times  it  is  contained  in  the 
solid.  The  process  is  essentially  comparison.  The  unit  cube 
is  a  cube  each  of  whose  edges  is  one  unit  long. 

152.  The  following  formulas  have  been  used : 

v  =  Iwh, 
v=s3. 

153.  The   product   of   factors   having   a   common    base 
equals  a  number  whose  base  is  the  same  as  the  factors 
and  whose  exponent  is  the  sum  of  the  exponents  of  the 
factors. 

154.  The   quotient   obtained   by   dividing   a  power  by 
another  power  having  the  same  base  is  a  number  whose 
base  is  the  common  base  of  the  given  powers  and  whose 
exponent  is  obtained  by  subtracting  the  exponent  of  the 
divisor  from  the  exponent  of  the  dividend. 


CHAPTER  VI 

THE  EQUATION  APPLIED  TO  FUNDAMENTAL 
ANGLE  RELATIONS 

155.  Fundamental  angle  relations.    In  Chapter  III  we 
discussed  the  different  kinds  of  angles  and  the  methods  of 
constructing  them.    In  this  chapter  we  shall  study  some 
of  the  fundamental  relations  between  angles  and  see  how 
the  equation  is  applied  to  them. 

156.  Relation  of  exterior  sides  of  supplementary  adjacent 
angles.    Draw  two  adjacent  angles  of  64°  and  116°,  of  75° 
and  105°,  of  157°  and  23°.    What  is  the  sum  of  each  pair? 
What  is  the  relation  of  the  exterior  sides  of  each  pair  ? 


E 


FIG.  93 


157.  Important  geometric  relation.  The  preceding  article 
illustrates  the  following  geometric  relation :  If  the  sum  of 
two  adjacent  angles  is  a  straight  angle,  their  exterior  sides 
form  a  straight  line. 

ill 


112  GENEKAL  MATHEMATICS 

EXERCISES 

1.  Show  that  the  geometric  relation  stated  in  Art.  157  agrees 
with  the  definition  of  a  straight  angle  (Art.  61). 

2.  In  Fig.  93  read  the  number  of  degrees  in  angles  XOA, 
XOB,  XOC,  XOY,  XOD,  XOE. 

3.  What  is  the  sum  of  /.XOA  and  Z.AOB? 

4.  Express  /.XOD  as  the  sum  of  two  angles. 

5.  Express  Z.AOB  as  the  difference  of  two  angles. 

6.  Express  /.XOE  as  the  sum  of  three  angles. 

158.  Sum  of  all  the  angles  about  a  point  on  one  side 
of  a  straight  line.    Draw  a  line  AB  and  choose  a  point 
P  on  it.    Draw  lines  Pit, 

PS,  and  PT  as  shown  in 
Fig.  94  and  find  the  sum 
of  the  four  angles  formed. 
Estimate  first  and  then 
measure  with  the  pro- 
tractor. What  seems  to 
be  the  correct  sum  ?  Ex- 
press the  sum  of  the  angles  x,  y,  z,  and  w  by  means  of 
an  equation.  Give  a  word  statement  for  the  equation. 

159.  Important  geometric  relation.    Art.  158  illustrates 
the  truth  of  the  geometric  relation  that  the  sum  of  all  the 
angles  about  a  point  on  one  side  of  a  straight  line  is  a  straight 
angle  (180°). 

EXERCISES 

1.  Find  the  value  of  x  and  the  size  of  each  angle  in  Fig.  95. 

2.  In  the  following  examples  each  expression  represents  one 
of  the  angles  into  which  all  the  angular  space  about  a  point 
on  one  side  of  a  straight  line  has  been  divided.    Write  an 
equation  expressing  the  sum  of  all  the  angles,  solve  for  x, 


FUNDAMENTAL  ANGLE   RELATIONS 


113 


and  find  the  size  of  each  angle  in  degrees.    Draw  figures  to 
illustrate  your  results  in  the  first  three  examples. 

(a)  x,  3  a-,  5  x  +  9. 

(b)  3x,  4z-10,  33-2*,  Wx  +  7. 

(c)  2  x  +  18},  5  x  +  91,  8J-  +  a-. 

(d)  2  (x  +  5),  3  x  +  24,  2  (35  +  x). 

(e)  5.83  x,  3,94  a-,  1.27  a-  +  11.55,  138.45  -  8.04  x. 


FIG.  95 


FIG.  96 


160.  Sum  of  all  the  angles  about  a  point  in  a  plane.  If 
we  choose  a  point  in  a  plane,  as  P  in  Fig.  96,  and  from  this 
point  draw  four  lines  so  as  to  make  four  angles,  we  can 
measure  these  angles  and  thus  determine  the  sum  of  all 
the  angles  in  a  plane  about  a  point. 


EXERCISES 

1.  Measure  the  angles  in  Fig.  96  and  write  an  equation 
expressing  their  sum. 

2.  What   is   another   way 
to  show  that  the  sum  of  the 
angles  that  exactly  fill  a  plane 
about  a  point  is  two  straight 
angles  (360°)?    See  the  defi- 
nition of  a  perigon  (Art.  61). 

3.  Find  the  value  of  x  and  the  size  of  each  angle  in  Fig.  97. 


3Z+80 
FIG.  97 


114  GENERAL  MATHKMATICS 

4.  The  expressions  in  the  following  examples  represent  the: 
angles  into  which  the  angular  space  about  a  point  in  a  plane 
has  been  divided.  Find  the  size  of  each  angle. 

(a)  3  x,  x,  2  x  +  35,  125  -  at. 

(b)  2x,  72  +  3x,  4*  -  10,  118. 

(c)  10  x  +  20-J-,  35|  -  x,  8  x  +  49. 

(d)  5  a-,  3x  +  27f,*7z  -  20,  9a-.  +  112j. 

(e)  x  +  1,  7  (a;  +  1),  3  (35  +  x),  2  x  +  169. 

(f)  3x,  117 +  15 a-,  9 a; -27. 

(g)  14  x  +  48,  28  x  +  106f ,  133^  -  6  x. 

The  first  two  exercises  in  this  article  show  that  the  sum 
of  all  the  angles  about  a  point  in  a  j)lafie  is  360°. 

161.  Left  side  of  an  angle;  right  side  of  an  angle.    If  in 
Fig.  98  we  imagine  ourselves   standing  at  the  vertex  of 
/.ABC  and  looking  off  over  the  angular  space,  say  in  the 
direction  BD,  then  the  side  BC  Q 

is  called  the  left  side  of  the  angle 

(because  it  lies  on  our  left),  and  -  a*    - 

the  side  BA  is  called  the  right  s\f&^~~' 

side  of  the  angle.  ^-''" 

B  Right  A 

162.  Notation.   In  lettering  an- 

,  ,  fi  .     .       .          f   .  FIG.  98 

gles  and  figures  it  is  often  desir- 
able to  denote  angles  or  lines  that  have  certain  characteristic 
likenesses  by  the  same  letter  so  as  to  identify  them  more 
easily.  It  is  clear  that  to  use  I  for  the  left  side  of  one 
angle  and  the  same  I  for  the  left  side  of  another  angle 
in  the  same  discussion  might  be  misleading.  In  order  to 
be  clear,  therefore,  we  let  ^  stand  for  the  left  side  of  one 
angle,  Z2  stand  for  the  left  side  of  a  second  angle,  and 
13  stand  for  the  left  side  of  a  third  angle,  etc.  Then  the 
three  sides  would  be  read  "  I  sub-one,"  "  I  sub-two,"  "  I  sub- 
three,"  etc. 


FUNDAMENTAL  ANGLE  RELATIONS 


115 


li 


FIG. 


163.  Important  geometric  relation.    Two  angles,  a;1  and 
xv  in  Fig.  99,  are  drawn  so  that  their  sides  are  parallel  left 
to  left  and  right  to 

right.  How  do  they 
seem  to  compare  in 
size  ?  Check  your 
estimate  by  measur- 
ing with  a  protractor. 
Give  an  argument 
showing  that  x-^  =  x¥ 
This  article  shows 

that  if  two  angles  have  their  sides  parallel  left  to  left  and  right 
to  right,  the  angles  are  equal. 

EXERCISE 

Draw  freehand  two  obtuse  angles  so  that  their  sides  will 
look  parallel  left  to  left  and  right  to  right.  (The  angles  should 
be  approximately  equal.  Are  they  ?) 

HINT.  Take  -two  points  for  vertices  and  in  each  case  imagine 
yourself  standing  at  the  point.  Draw  the  left  sides  to  your  left  and 
the  right  sides  to  your  right.  Assume  the  drawing  correct  and  prove 
the  angles  equal. 

164.  Important  geometric  relation.    Two  angles,  x  and 
y,  in  Fig.  100,  have  been  drawn  so  that  their  sides  are  par- 
allel left  to  right  and  right  to  left.    What  relation  seems 
to  exist  between  them  ?  Meas- 
ure   each    with    a   protractor. 

Give    an    argument    showing 
that  z  +  #  =  180°.  • 

This  article  shows  that  if 
two  angles  have  their  sides  par- 
allel left  to  right  and  right  to 
left,  their  sum  is  a  straight  angle.  FIG.  100 


1 1 


116  GENERAL  MATHEMATICS 

EXERCISE 

Practice  drawing  freehand  a  pair  of  angles  whose  sides  are 
parallel  according  to  the  conditions  in  the  theorem  of  Art.  164. 
Is  the  sum  approximately  180°  ? 

165.  Supplementary  angles ;  supplement.  Two  angles 
whose  sum  is  equal  to  a  straight  angle  (180°)  are  said 
to  be  supplementary  angles.  Each 
angle  is  called  the  supplement  of 
the  other. 


166.  Supplementary      adjacent  F      101 

angles.  Place  two  supplementary 

angles  adjacent  to  each  other  as  in  Fig.  101.    Angles  so 
placed  are  called  supplementary  adjacent  angles. 

EXERCISES 

1.  In  Fig.  101  what  is  the  angle  whose  supplement  is  Zee? 

2.  In  Fig.  102  are  several  angles,  some  pairs  of  which  are 
supplementary.    Make  tracings  of  these  angles  on  paper  and 
by  placing  them  adjacent  decide  which  pairs  are  supplementary. 


FIG.  102 

3.  State  whether  the   following  pairs    of   angles  are  sup- 
plementary :  40°  and  140° ;  30°  and  150° ;  35°  and  135° ;  55° 
and  135°. 

4.  How  many  degrees  are  there  in  the  supplement  of  an 

angle  of  30°  ?  of  90°  ?  of  150°  ?  of  x°  ? 

2  s° 

5.  What  is  the  supplement  of  y°  ?  of  z°  ?  of  3  w°  ?  of  —  ? 

o 


117 

6.  Write  the  equation  which  expresses  the  fact  that  y°  and 
130°  are  supplementary  and  solve  for  the  value  of  y. 

7.  Write  equations  that  will  show  that  each  of  the  follow- 
ing pairs  of  angles  are  supplementary  : 

(a)  y°  and  80°.  (d)  30°  and  y°  +  40°. 

(.b)  90°  and  z°.  (e)  3  y°  +  5°  and  12  y°  -  4°. 

(c)  x°  and  y°.  (f)  f  x°  and  l£x°  +  75  J°. 

8.  Two  supplementary  angles  have  the  values  2x°  +  25° 
and  x°  +  4°.    Find  x  and  the  size  of  each  angle. 

9.  What  is  the  size  of  each  of  two  supplementary  angles 
if  one  is  76°  larger  than  the  other  ? 

10.  One  of  two  supplementary  angles  is  33°  smaller  than  the 
other.  Find  the  number  of  degrees  in  each. 

11..  What  is  the  number  of  degrees  in  each  of  two  supple- 
mentary angles  whose  difference  is  95°  ? 

12.  Find  the  value  of  x  and  the  angles  in  the  following 
supplementary  pairs : 

(a)  x°  and  6  x°. 

(b)  2  x°  and  3  x°  +  2°. 

(c)  4£  x°  and  6  x°. 

(d)  2  aj°  +  5°  and  7  x°  -  8°. 

13.  Write  the  following  expressions  in  algebraic  language : 

(a)  Twice  an  angle  y. 

(b)  Four  times  an  angle,  plus  17°. 

(c)  23°  added  to  double  an  angle. 

(d)  Seven  times  an  angle,  minus  14°. 

(e)  45°  less  than  an  angle. 

(f )  52°  subtracted  from  four  times  an  angle. 

(g)  Twice  the  sum.  of  an  angle  and  10°. 
(h)  One  half  the  difference  of  22°  and  x°. 

14.  If  an  angle  is  added  to  one  half  its  supplement,  the 
sum  is  100°.    Find  the  supplementary  angles. 


118  GENERAL  MATHEMATICS 

15.  If  an  angle  is  increased  by  5°  and  if  one  fourth  of  its 
supplement  is  increased  by  25°,  the  sum  of  the  angles  thus 
obtained  is  90°.    Find  the  supplementary  angles. 

16.  Construct  the  supplement  of  a  given  angle. 

17.  Find  the  size  of  each  of  the  following  adjacent  pairs  of 
supplementary  angles  : 


(a)  0»llft:+«i  (d)         -60,  130-     *. 

• 

(b)  |  x  +  32,  88  -  \x.  (e)  2  (*  +  10),  ^±« 

(c)  |  +  150,  I  -  10.  (f)  65  +  2-f,  92  +  ^. 

167.  Construction  problem.  To  construct  the  supplements 
of  two  equal  angles. 

Construction.  Let  a;  and 
y  be  the  given  angles. 
Construct  Z  z,  the  supple- 

ment of  Zz,  adjacent  to  z  /x  w  /y 

it  (Fig.  103).   In  the  same 
manner  construct  Zw-,  the    Fm-  103'    HOWTO  CONSTRUCT  THE  SCPPLK- 

,     .  MENTS    OF    T\VO    GlVEX    AHOLES 

supplement  of  Z.y. 

Compare  the  supplements  of  Z  x  and  Z  y  and  show  that 

/-Z=/-W. 

This  article  shows  that  the  Supplements  of  equal  angles 
are  equal. 

EXERCISES 

1.  Prove  the  preceding  fact  by  an  algebraic  method. 

HINT.  In  Fig.103  prove  that  if  Zx  +  Zz  =180°  and  Z.y  +  Zro  =180°, 
then  Zz  =  Zw. 

2.  Are  supplements  of  the  same  angle  equal  ?    Why  ? 


FUNDAMENTAL  ANGLE  RELATIONS 


119 


3.  Show  that  the  bisectors  of  two  supplementary  adjacent 
angles  are  perpendicular  to  each  other ;  for  example,  in  Fig.  104 
show  that  Z  x  +  Z.  y  =  90°. 

4.  In  Fig.  104,  if  Z.  BOD  =  60° 
and  Z.  AOD  =  120°,  find  the  size 
of  Z.  x  and  Z.  y. 

*5.  The  following  examples 
furnish  a  review.  In  each  case 
solve  for  the  value  of  the  un- 
known, and  check. 


(»)  Y 

(b)  \y 


OD 


(e) 


t    ^ 

(g) 


.    "         19 

h  <  =  12. 


16  +  -*  =  3. 
o 


168.  Complementary  angles.  If  the 
sum  of  two  angles  is  a  right  angle,  the 
two  angles  are  called  complementary 
angles.  Each  angle  is  called  the  com- 
plement of  the  other.  Thus,  in  Fig.  105 
Z.x  is  the  complement  of  Z  y. 


FIG.  105 


EXERCISES 

1.  What  is  the  complement  of  30°  ?  of  60°  ? 

2.  Are  23°  and  57°  complementary  ?  32°  and  58°  ? 

3.  Draw  two   complementary  angles   of  40°  and  50°  and 
place  them  adjacent.    Check  the  construction. 

4.  What  is  the  relation  existing  between  the  exterior  sides 
of  two  adjacent  complementary  angles  ? 


120 


GENERAL  MATHEMATICS 


5.  In  Fig.  106  decide  by  means  of  tracing  paper  which 
pairs  of  angles  seem  to  be  complementary. 

6.  What  are  the   complements  of  the  following   angles : 
20°?  50°?  12£°?  48|°?  x°?  3y°?  ^? 

7.  40°  is  the  complement  of  y°.    How  many  degrees  does 
y  represent  ? 

8.  Write  the  equation  which  says  in  algebraic  language 
that  x°  and  50°  are  complementary  and  solve  for  the  value  of  x. 


FIG.  106 

9.  In  the  equation  x°  +  y°  =  90°  is  there  more  than  one 
possible  pair  of  values  of  x  and  y  ?  Explain. 

10.  Write  equations  that  will  express  the   fact   that  the 
following  pairs  of  angles  are  complementary : 

(a)  x°  and  40°.  (c)  x°  +  25°  and  x°  -  30°. 

(b)  35°  and  y°.  (d)  2  x°  -  3°  and  3  x°  +  8°. 

11.  Write  the  following  expressions  in  algebraic  language: 

(a)  The  sum  of  angle  x  and  angle  y. 

(b)  Four  times  an  angle,  increased  by  15°. 

(c)  85°  diminished  by  two  times  an  angle. 

(d)  Five  times  the  sum  of  an  angle  and  13°. 

(e)  Three  times  the  difference  between  an  angle  and  12°. 

(f)  Four  times  an  angle,  minus  6°. 


FUNDAMENTAL  ANGLE  RELATIONS         121 


12.  Find  two  complementary  angles  such  that  one  is  32° 
larger  than  the  other. 

13.  Find  two  complementary  angles  such  that  one  is  41° 
smaller  than  the  other. 

14.  Find  the  number  of  degrees  in  the  angle  x  if  it  is  the 
complement  of  3  x° ;  of  5  x° ;  of  8^  x°- 

15.  Find  the  number  of  degrees  in  the  angle  x  if  it  is 
the  complement  of  twice  itself;  of  five  times  itself;  of  one 
third  itself. 

16.  Construct  the  complement  of  a  given  acute  angle. 

169.  Construction  problem.    To   construct   the    comple- 
ments of  two  equal  acute  angles. 

Construction.    Let  Zx  and  /.y  be  the  given  angles.    Draw  the 
complement  of  £x  adjacent  to  it  (Fig.  107).    Do  the  same  for  £y. 


FlG.  107.    HOW  TO   CONSTRUCT  THE  COMPLEMENTS  OF.  TWO  GlVEN  ANGLES 

Compare  the  complements  of  Ax  and  y  and  show  that 
Z0  =  /.w. 

This  construction  problem  shows  that  complements  of 
equal  angles  are  equal. 

EXERCISES 

1.  Prove  the  preceding  relation  by  an  algebraic  method. 

2.  Does  it  follow  that  complements  of  the  same  angle  are 
equal?    Why? 


122 


GENERAL  MATHEMATICS 


170.  Vertical  angles.  Dra\v  two  in- 
tersecting straight  lines  AB  and  CD  as 
in  Fig.  108.  The  angles  x  and  z  are 
called  vertical,  or  opposite,  angles.  Note 
that  vertical  angles  have  a  common 
vertex  and  that  their  sides  lie  in  the 
same  straight  line  but  in  opposite  direc- 
tions. Thus,  vertical  angles  are  angles 
which  have  a  common  vertex  and  their 
sides  Jying  in  the  same  straight  line  -A  D 

but  in   opposite  directions.    Are  w  and    FIG.  108.  VERTICAL 
y  in  Fig.  109  vertical  angles? 


ANGLES 


EXERCISES 
(Exs.  1-6  refer  to  Fig.  108) 

1.  Make  a  tracing  of  /-  x  and  /-  z  and  compare  them  as  to  size. 

2.  Check  your  estimate  in  Ex.  1  by  measuring  the  two  angles 
with  a  protractor. 

3.  What  is  the  sum  of  /-  x  and  /.  y  ?  of  Z.  z  and  /-  y  ? 

4.  Show  that  x  +  y  =  z  -j-  y. 

5.  How  does  Ex.  4  help  in  obtaining  the  relation  between 
x  and  z  ?    What  is  this  relation  ? 

6.  Show  that  y  +  x  =  x  -+-  w  and  from  this  that  y  =  w. 

The  six  exercises  above  show  that  if  two  lines  intersect, 
the  vertical  angles  are  equal. 

171.  Value  of  mathematical  thinking.  The  preceding 
relation  between  vertical  angles  is  of  course  so  easily  seen 
that  in  most  cases  the  truth  would  be  granted  even  with- 
out measuring  the  angles  involved.  However,  the  discus- 
sion in  Exs.  3-6  above  is  another  simple  illustration  of 


123 

the  power  of  mathematical  thinking  which  makes  the  dis- 
covery of  new  truths  rest  finally  on  nonmeasurement,  that 
is  to  say,  on  an  intellectual  basis.  This  type  of  thinking 
will  be  used  to  an  increasing  extent  in  subsequent  work. 

EXERCISES 

1.  Upon    what    does    the   proof    (Exs.  3-6,   Art.   170)    of 
the   geometric   relation   concerning   vertical   angles   rest  ? 

2.  Find  x  and  the  size  of 

each  angle  in  Fig.  109.  — ~^_^__  o 

First    method.    Since    vertical 
angles  are  equal, 


Then  3r  +  4  =  2ar  +  10. 

Subtracting  4  from  each  member, 

3x  =  2x  +  6. 

Subtracting  2  x  from  each  member, 

*  =  «. 

Substituting  6  for  x,       3  x  +  4  =  3  •  6  +  4  =  22, 
2  x  +  10  =  2  .  6  +  10  =  22, 
9  x  +  104  =  9  .  6  +  104  =  158  (for  Z.  BOC), 
and  since  vertical  angles  are  equal, 


Check.    22  +  22  +  158  +  158  =  360°. 

Second  method.    By  definition  of  supplementary  angles, 

8*  +  4  +  9*  +  104  =180. 
Solving,         .  r  =  6. 

The  remainder  of  the  work  is  the  same  as  that  of  the  first  method. 


124  GENERAL  MATHEMATICS 

3.  Find  the  values  of  the  unknowns  and  each  of  the  follow- 
ing vertical  angles  made  by  two  intersecting  straight  lines  : 

(a)  3  x  +  15  and  5x  —  5.  (f)    -J  x  +  $  x  and  f  x  +  55. 

(b)  ,  +  105  and  IS*  +  15.  £*         5, 

(c)  cc.—  10  and  2  a;  —  160.  4  2 

(d)  +  181  and       +  21£.       (h)      +     and     +  18. 


Q  ~  8  T          ^  ->•  9  T 

(e)|*-8andj  +  12.  (i)  ^  -  -  ^  and  ^  +  11  ]. 

172.  Alternate-interior  angles.  In  Fig.  110  the  angles 
x  and  y,  formed  by  the  lines  AB,  CD,  and  the  transversal 
EF,  are  called  alternate-interior  angles  (on  alternate  sides 
of  EF  and  interior  with  respect  to  AB  and  CD~). 


FIG.  110  E  FIG.  Ill 


EXERCISES 

(Exs.  1-4  refer  to  Fig.  Ill) 

1 .  Measure  and  compare  Z  x  and  Z  //. 

2.  The  lines  AB  and  CD  and  FE  are  drawn  so  that  Z.x=Z.y. 
What  seems  to  be  the  relation  between  the  lines  AB  and  CD? 

3.  Show  that  if  Z  x  =  Z  y,  then  Z  y  =  Z  z. 

4.  Show  that  4-^  is  parallel  to  CD  (see  the  definition  for 
parallel  lines,  Art.  87). 


FUNDAMENTAL  ANGLE  RELATIONS 


125 


Exercises  1-4  show  that  if  the  alternate-interior  angles 
formed  by  two  lines  and  a  transversal  are  equal,  the  lines  are 
parallel. 

The  proof  may  take  the  following  brief  form: 

Proof.  In  Fig.  Ill,  Z  x  =  Z  y  (given).  Z  x  =  Z  z  (vertical  angles  are 
equal).  Then  Z?/  =  Z.z  (things  equal  to  the  same  thing  are  equal  to 
each  other).  Therefore  AB  II  CD  (by  definition  of  II  lines,  Art.  87). 

EXERCISE 

In  Fig.  112  construct  a  line  through  P  parallel  to  the  line 
AB  by  making  an  alternate-interior  angle  equal  to  /.x.  Show 
why  the  lines  are  parallel.  F 

A \ B 


FIG.  112 

173.  Interior  angles  on  the  same  side  of  the  transversal. 
In  Fig.  113  angles  x  and  y  are  called  interior  angles  on  the 
same  side  of  the  transversal. 

EXERCISES 

1.  Measure  angles  x  and  y  in  Fig.  113  and  find  their  sum. 

2.  In  Fig.  114  the  lines  are  drawn  so  that  /Lx  +  Z.y  =  180°. 


What  seems  to  be  the  relation 
between  AB  and  CD? 

3.  Prove  that  if  the  interior 
angles  on  the  same  side  of  a, 
transversal  between  two  par- 
allel lines  are  supplementary, 
the  lines  are  parallel. 


'\- 


-D 


FIG.  114 


126  GENERAL  MATHEMATICS 

4.  In  Fig.  115  select  all  the  pairs  of  corresponding  angles, 
alternate-interior  angles,  and   in-  F 

terior  angles  on  the  same  side  of  / 

x  /?/ 
the  transversal.  „./„'        B 

174.  Important  theorems  relat- 
ing to  parallel  lines.  The  follow-  c y^ D 

ing  exercises  include  theorems  / 

which  supplement  the  work  of  'E 

Arts.  172  and  173.  FIG.  115  • 

EXERCISES 

1.  Show  by  reference  to  the  definition  of  parallel  lines  in 
Art.  87  that  if  two  parallel  lines  are  cut  by  a  transversal,  the 
corresponding  angles  are  equal. 

2.  Show  that  if  two  parallel  lines  are  cut  by  a  transversal, 
the  alternate-interior  angles  are  equal. 

3.  Show  that  if  two  parallel  lines  are  cut  by  a  transversal, 
the  interior  angles  on  the  same  side  of  the  transversal  are 
supplementary. 

4.  Two  parallel  lines  are  cut   by  a  transversal  so  as  to 
form  angles  as  shown  in  Fig.  116.    Find  x  and  the  size  of 
all  the  eight  angles  in  the  figure. 

V       Ab  <K'^ 


\  X 

FIG. 116  FIG.  117 

5.  Find  x  and  all  the  eight  angles  in  Fig.  117. 

6.  Draw  two  parallel  lines  and  a  transversal.    Select  all  the 
equal  pairs  of  angles ;  all  the  supplementary  pairs. 


FUNDAMENTAL  ANGLE  RELATIONS         1.27. 

175.  Outline  of  angle  pairs  formed  by  two  lines  cut  by 
a  transversal.    When  two  lines  are  cut  by  a  transversal, 

as  in  Fig.  118, 

"a  and  e~] 

the  angles  of  the    b  and  / 

7       IT?-  are  called  corresponding  angles ; 
angle  pairs          a  and  h 

c  and  g\ 

angles  c,  d,  e,  f  are  called  interior  angles ; 
angles  a,  5,  #,  h  are  called  exterior  angles ; 

the  angles  of  the  f  d  and  e  1  are  called  interior  angles  on  the 
angle  parrs       j  c  and  /j       same  side  of.  the  transversal ; 

*:',,     f  ,        ,    J   on  opposite  sides  of  the  trans- 
the  angles  of  the  \  a  and  / 

,      >•       versal  are  called  alternate- 
angle  pairs          c  and  e\ 

J        interior  angles  ; 

,  j "]  on  opposite  sides  of  the  trans- 

the  angles  ot  the    b  and  h\                                    ,  .   , 

^           ,  Y       versal  are  called   alternate- 

angle  pairs          a  and  a  \ 

[  J        exterior  angles. 


FIG.  119 
The  student  should  remember 

(a)  that  corresponding  angles  are  equal, 

(b)  that  alternate-interior  angles  are  equal, 

(c)  that  alternate-exterior  angles  are  equal, 

(d)  that  interior  angles  on  the  same  side  of 

the  transversal  are  supplementary, 
only  when  the  lines  cut  by  the  transversal  are  parallel  (Fig.  119). 


128  GENERAL  MATHEMATICS 

SUMMARY 

176.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases:  left  side  of  an  angle,  right  side 
of  an  angle^  parallel  right  to  right  and  left  to  left,  par- 
allel right  to  left  and  left  to  right,  supplementary  angles, 
supplement,  supplementary-adjacent  angles,  complementary 
angles,  complement,  vertical  angles,  alternate-interior  angles, 
interior  angles  on  the  same  side  of  the  transversal. 

177.  The  following  fundamental  constructions  have  been 
presented : 

1.  How  to  construct  the  supplement  of  a  given  angle. 

2.  How  to  construct  the  supplements  of  two  equal  given 
angles. 

3.  How  to  construct  the  complement  of  -a  given  angle. 

4.  How  to  construct  the    complements  of   two   equal 
angles. 

5.  A  new  method  of  drawing  parallel  lines. 

6.  How  to  form  vertical  angles. 

178.  This  chapter  has  discussed  the  following  funda- 
mental geometric  relations: 

1.  If  the  sum  of  two  adjacent  angles  is  a  straight  angle, 
their  exterior  sides  form  a  straight  line. 

2.  The  sum  of  all  the  angles  about  a  point  on  one  side 
of  a  straight  line  is  a  straight  angle  (180°). 

3.  The  sum  of  all  the  angles  in  a  plane  about  a  point 
is  two  straight  angles  (360°). 

4.  If  two  angles  have  their  sides  parallel  left  to  left  and 
right  to  right,  the  angles  are  equal. 

5.  If  two  angles  have  their  sides  parallel  left  to  right 
and  right  to  left,  the  angles  are  supplementary. 

6.  Supplements  of  the  same  angle  or  of  equal  angles 
are  equal. 


FUNDAMENTAL  ANGLE  RELATIONS          129 

7.  Complements  of  the  same  angle  or  of  equal  angles 
are  equal. 

8.  If  two  lines   intersect,  the  vertical  angles   formed 
are  equal. 

9.  Two  lines  cut  by  a  transversal  are  parallel 

(a)  if  the  corresponding  angles  are  equal ; 

(b)  if  the  alternate-interior  angles  are  equal; 

(c)  if  the  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplementary. 

10.  If  two  parallel  lines  are  cut  by  a  transversal,  then 

(a)  the  corresponding  angles  are  equal ; 

(b)  the  alternate-interior  angles  are  equal ; 

(c)  the  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplementary. 


CHAPTER  VII 


THE  EQUATION  APPLIED  TO  THE  TRIANGLE 

179.  Notation  for  triangles.     It  is  customary  to  denote 
the  three  points  of  intersection  of  the  sides  of  a  triangle 
by  capital  letters  and  the  three  sides  which  He  opposite 
these  respective  sides  with  the  corresponding  small  letters. 
Thus,  in  Fig.  120  we  denote 

the  points  of  intersection 
of  the  sides  (the  vertices) 
by  A,  B,  and  C,  and  the 
sides  opposite  by  a,  6,  and  c. 
The  sides  may  also  be 

read  BC,  AC,  arid  AB.  The  symbol  for  "  triangle"  is  a  small 
triangle  (A).  The  expression  A  ABC  is  read  "  triangle  ABC." 
The  three  angles  shown  in  Fig.  120  are  called  interior  angles. 

180.  Measuring  the  interior  angles  of  a  triangle.  We  shall 
now  consider  some  of  the  methods  of  measuring  the  interior 
angles  of  a  triangle. 

EXERCISES 


ANGLE 

TRIANGLE  ABC 
No.  OF  DEGREES 

TRIANGLE  DEF 
No.  OF  DEGREES 

TRIANGLE  GHI 
No.  OF  DEGREES 

Estimated 

Measured 

Estimated 

Measured 

Estimated 

Measured 

X 

y 

z 

Sum 

130 


131 


1.  Fill  in  the  table  on  the  preceding  page  with  reference 
to  the  triangles  ABC,  DEF,  and  GHI  (Fig.  121). 


FIG.  121 

2.  Draw  a  triangle  on  paper  (Fig.  122).  Cut  it  out  and 
tear  off  the  corners  as  shown  in  Fig.  123.  Then  place  the 
three  angles  adjacent  as  shown.  What 
seems  to  be  the  sum  of  the  three  angles 
of  the  triangle  ?  Test  your  answer  with 
a  straightedge. 

181.  Theorem.  The  results  of  Exs.  1  FIG.  122 
and  2,  above,  illustrate  the  geometric 

relation  that  the  sum  of  the  interior 
angles  of  a  triangle  is  a  straight  angle 
(180°).  The  statement  "The  sum 
of  the  interior  angles  of  a  triangle  is 
a  straight  angle"  can  be  proved  by  more  advanced  geo- 
metric methods.  Such  a  statement  of  a  geometric  relation 
to  be  proved  is  called  a  theorem. 

182.  More   advanced   methods  of 
proof  for  the  preceding  theorem.   The 
truth  of  the  theorem,  that  the  sum  of 
the  interior  angles  of  a  triangle  is    Ji 
180°  may  be  illustrated  as  follows: 
Draw  a  triangle  as  in  Fig.  124.  Place 

a  pencil  at  A  as  indicated  in  the  figure,  noting  the  direction 
in  which  it  points.    Rotate  the  pencil  through  angle  A  as 


FIG.  123 


FIG.  124 


132 


GENERAL  .M  ATM  K.MATK  'S 


sliown  by  the  arrowhead.  Then  slide  it  along  AB  to  the 
position  indicated  in  the  figure.  Rotate  the  pencil  next 
through  angle  B  as  indicated  and  slide  it  along  BC  to 
the  position  sliown.  Then  rotate  the  pencil  through  angle 
C  to  the  last  position  shown.  This  rotation  through 
angles  .1.  /;,  and  C  leaves  the  point  of  the  pencil  in  what 
position  in  respect  to  its  original  position  ?  What  part  of 
a  complete  turn  has  it  made  ?  Through  how  many  right 
angles  has  it  turned  ?  Through 
how  many  straight  angles  ? 
Through  how  many  degrees? 

The  theorem  that ':  the  sum 
of  the  interior  angles  of  a  tri- 
angle  is  180°  "  may  be  proved 
as  follows : 

(oven  triangle  AB<'(Y\%.  12o),  to  prove  that  Z.I  +  Z/>  +  ZC'  =  180°. 

Proof 


STATE.MK.NT> 


REASONS 


Draw 


Then 


And 


Z  if  -  Z  B. 


But  Z ./.;  4-  Z  //  +  A  :  =  180° 


.-.  z.i 


Because  corresponding  angles 
formed  by  two  parallel  lines  cut 
by  a  transversal  are  equal. 

Because  alternate-interior  an- 
gles formed  by  two  parallel  lines 
cut  by  a  transversal  are  equal. 

Because  the  sum  of  all  the 
angles  about  a  ]  >oiut  in  a  plane  on 
one  side  of  a  straight  line  is  180°. 

By  substituting  Z.A  for  Zr. 
i  Z  B  for  Z  if.  and  Z  C  for  Z  r. 


This  is  n  more  formal  proof  of  the  theorem,  inasmuch  as  it 
is  independent  of  measurement.  Write  an  equation  which  will 
express  the  number  of  degrees  in  the  sum  of  the  angles  of 
a  triangle. 


.  EQUATION  APPLIED  TO  THE  TRIANGLE     133 

This  equation  is  a  very  useful  one,  as  it  enables  us  to 
find  one  angle  of  a  triangle  when  the  other  two  are  known. 
Thus,  if  we  know  that  two  angles  of  a  triangle  are  50° 
and  70°,  we  know  that  60°  is  the  third  angle.  This  is  of 
great  practical  value  to  the  surveyor,  who  is  thus  enabled 
to  know  the  size  of  all  three  angles  of  a  triangle  by 
measuring  only  two  directly. 

HISTORICAL  XOTK.  Thales  (040  B.C.  -  about  o50  B.C.),  the  founder 
of  the  earliest  Greek  school  of  mathematics,  is  supposed  to  have  known 
that  the  sum  of  the  angles  of  a  triangle  is  two  right  triangles. 
Someone  has  suggested  that  this  knowledge  concerning  the  sum 
of  the  angles  of  a  triangle  may  have  been  experimentally  demon- 
strated by  the  shape  of  the  tiles  used  in  paving  floors  in  Th-ales' 
day.  What  has  been  regarded  as  the  most  remarkable  geometrical 
advancement  of  Thales  was  the  proof  of  a  theorem  which  depended 
upon  the  knowledge  that  the  sum  of  the  angles  of  a  triangle  is  two 
right  angles.  It  is  related  that  when  Thales  had  succeeded  in  proving 
the  theorem,  he  sacrificed  an  ox  to  the  immortal  gods.  The  large 
number  of  stories  told  about  Thales  indicates  that  he  must  have 
been  a  man  of  remarkable  influence  and  shrewdness  both  in  science 
and  in  business.  Thus,  we  read  that  at  one  time  he  cornered  the  olive 
market  and  that  at  another  time  he  was  employed  as  engineer  to 
direct  a  river  so  that  a  ford  might  be  constructed.  The  following 
story  is  told  illustrative  of  his  shrewdness : 

It  is  said  that  once  when  transporting  some  salt  which  was  loaded 
on  mules  one  of  the  animals,  slipping  in  a  stream,  got  its  load  wet  and  so 
caused  some  of  the  salt  to  be  dissolved.  Finding  its  burden  thus  light- 
ened, it  rolled  over  at  the  next  ford  to  which  it  came  ;  to  break  it  of 
this  trick,  Thales  loaded  it  with  rags  and  sponges,  which,  by  absorbing 
the  water,  made  the  load  heavier  and  soon  effectually  cured  it  of  its 
troublesome  habit.1 

183.  Problems  involving  the  theorem  "  The  sum  of  the 
interior  angles  of  a  triangle  is  a  straight  angle."  In 
the  problems  that  follow  the  pupil  will  need  to  apply  the 
theorem  proved  in  the  preceding  article. 

1  Ball,  "A  Short  Account  of  the  History  of  Mathematics,"  p.  14. 


134  GENERAL  MATHEMATICS 

EXERCISES 

In  the  following  problems 

(a)  Draw  freehand  the  triangle. 

(b)  Denote  the  angles  properly  as  given. 

(c)  Using  the  theorem  of  Art.  182,  write  down  the  equation 
representing  the  conditions  of  the  problem. 

(d)  Solve  the  equation  and  find  the  value  of  each  angle. 

(e)  Check  your  solution" by  the  conditions  of  the  problem. 

1.  The  angles  of  a  triangle  are  x,  2x,  and  3x.    Find  x  and 
the  number  of  degrees  in  each  angle. 

2.  The  first  angle  of  a  triangle  is  twice  the  second,  and  the 
third  is  three  times  the  first.    Find  the  number  of  degrees  in 
each  angle. 

3.  If  the  three  angles  of  a  triangle  are  equal,  what  is  the 
size  of  each  ? 

4.  If  two  angles  of  a  triangle  are  each  equal  to  30°,  what 
is  the  value  of  the  third  angle  ? 

5.  One  angle  of  a  triangle  is  25°.    The  second  angle  is  55° 
larger  than  the  third.    How  large  is  each  angle  ? 

6.  The  first  angle  of  a  triangle  is  four  times  the  second, 
and  the  third  is  one  half  the  first.    Find  e.ach  angle. 

7.  Find  the  angles  of  a  triangle   if  the  first  is  one  half 
of  the  second,  and  the  third  is  one  third  of  the  first. 

8.  The  first  angle  of  a  triangle  is  two  fifths  as  large  as 
another.    The  third  is  four  times  as  large  as  the  first.    How 
large  is  each  angle  ? 

9.  Find  the  angles  of  a  triangle  if  the  first  angle  is  16°  more 
than  the  second,  and  the  third  is  14°  more  than  the  second. 

10.  Find  the  angles  of  a  triangle  if  the  difference  between 
two  angles  is  15°,  and  the  third  angle  is  43°.. 

11.  The  first  angle  of  a  triangle  is  30°  more  than  the  second, 
and  the  third  is  two  times  the  first.    Find  the  angles. 


EQUATION  APPLIED  TO  THE  TRIANGLE     135 

12.  Find  the  angles  of  a  triangle  if  the  first  angle  is  twice 
the  second,  and  the  third  is  15°  less  than  two  times  the  first. 

13.  The  angles  of  a  triangle  are  to  each  other  as  1,  2,  3. 
What  is  the  size  of  each  ? 

HINT.    Let  x  =  the  first,  2  x  the  second,  and  3  x  the  third. 

14.  Find  the  angles  of  a  triangle  if  the  first  is  2^  times  the 
second  increased  by  10°,  and  the  third  is  one  fourth  of  the 
second. 

15.  In  a  triangle  one  angle  is  a  right  angle;  the  other  two 

%K> 

angles  are  represented  by  x  and  -  respectively.   Find  each  angle. 

16.  How  many  right  angles   may  a  triangle  have  ?    How 
many  obtuse  angles  ?    How  many  acute  angles  at  most  ?    How 
many  acute  angles  at  least  ? 

17.  Two  angles  x  and  y  of  one  triangle  are  equal  respec- 
tively to  two  angles  m  and  n  of  another  triangle.    Show  that 
the  third  angle  of  the  first  triangle  equals  the  third  angle  of 
the  second  triangle. 

184.  Theorem.   By  solving  Ex.  17  we  obtain  the  theorem 
If  two  angles  of  one  triangle  are  equal  respectively  to  two 
angles*  of  another  triangle,  the  third  angle  of  the  first  is  equal 
to  the  third  angle  of  the  second. 

185.  Right  triangle.    If  one  angle  of  a  triangle  is  a  right 
angle,  the  triangle  is  called  a  right  triangle.    The  symbol 
for  "  right  triangle  "  is  rt.  A. 

EXERCISES 

1.  Show  that  the  sum  of  the  acute  angles  of  a  right  triangle 
is  equal  to  a  right  angle. 

2.  Find  the  values  of  the  acute  angles  of  a  right  triangle  if 
one  angle  is  two  times  the  other ;  if  one  is  5°  more  than  three 
times  the  other. 


130 


GENEKAL  MATHEMATICS 


/'  /' 

3.  The  acute  angles  of  a  right  triangle  are  ^  and  r  • 


Find 


and  the  number  of  degrees  in  each  angle. 

4.  Draw  a  right  triangle  on  cardboard  so  that  the  two  acute 
angles  of  the  triangle  will  contain  30°  and  60°  respectively. 
Use  the  protractor. 

HINT.  First  draw  a  right  angle.  Then  at  any  convenient  point 
in  one  side  of  the  right  angle  construct  an  angle  of  60°  and  produce 
its  side  till  a  triangle  is  formed.  Why  does  the  third  angle  equal.30°? 

*5.  Cut  out  the  cardboard  triangle  made  in  Ex.  4  and  tell 
what  angles  may  be  constructed  by  its  use  without  a  protractor 
or  tracing  paper. 

6.  Draw  on  cardboard  a  right  triangle  whose  acute  angles 
are  each  equal  to  45°,  cut  it  out,  and  show  how  it  may  be  used 
to  draw  angles  of  45°  and  90°  respectively. 

186.  Wooden  triangles.    A  wooden  triangle  is  a  triangle 
(usually   a  right  one)  made  for  convenience  in  drawing 
triangles  on  paper  or  on  the  blackboard  (see  Fig.  126). 
The  acute  angles  are  usually  60°  and  30 

or  45°  and  45°.  These 
wooden  right  triangles  fur- 
nish a  practical  method  of 
drawing  a  perpendicular  to 
a  line  at  a  given  point  on 
that  line.  If  no  triangles 
of  this  kind  can  be  had, 
a  cardboard  with  two  per- 
pendicular edges  or  a  card- 
board right  triangle  will  serve  the  purpose  just  as  well. 

187.  Set  square.    A  set  square  is  made  up  of  a  wooden 
triangle  fastened  to  a  straightedge  so  that  it  will  slide 
along  the  straightedge  (see  Fig.  127). 


FIG.  126.    WOODEN 
TRIANGLE 


FIG.  127.    SET 
SQUARE 


EQUATION  APPLIED  TO  THE  TRIANGLE     137 

EXERCISES 

1.  Construct  a  right  triangle,  using  the  method  of  Art.  80. 

2.  The  set  square  (Fig.  127)  is  a  mechanical  device  for  drawing 
parallel  lines.    Show  how  it  may  be  used  to  draw  parallel  lines. 

3.  Show  that  two  lines  perpendicular  to  the  same  line  are 
parallel  (see  the  definition  of  parallel  lines,  Art.  87). 

4.  Show   how  a  wooden   triangle   may  be   used   to   draw 
parallel  lines. 

5.  What  are  three  ways  of  drawing  parallel  lines? 

188.  Problems  concerning  the  acute  angles  of  a  right 
triangle.  The  following  problems  will  help  the  student  to 
understand  the  relations  concerning  the  acute  angles  of 
a  right  triangle. 

EXERCISES 

1.  Draw  a  right  triangle  as  in  Fig.  128  and  show  that  angles 
.  I  and  B  are  complementary ;  that  is,  show  that  /.  A  +  /.B  =  90°. 

NOTE.  In  lettering  a  right  triangle 
ABC  the  letter  C  is  usually  put  at  the 
vertex  of  the  right  angle. 

2.  State   a   theorem    concerning 
the  acute  angles  of  a  right-angled 
triangle. 

3.  If  Z.I  =Z B  (Fig.  128),  what 

is  the  size  of  each  ?    How  do  the  sides  about  the  right  angle 
compare  in  length  ? 

4.  If  /.A  is  twice  as  large  as  /.B  (Fig.  128),  what  is  the  size 
of  each  ? 

*5.  Using  the  method  of  Ex.  4,  Art.  185,  draw  a  right  triangle 
whose  acute  angles  are  30°  and  60°  respectively.  Measure 
the  side  opposite  the  30-degree  angle.  Measure  the  side  oppo- 
site the  90-degree  angle.  (This  side  is  called  the  hypotenuse.) 
Compare  the  two  results  obtained. 


138 


GENERAL  MATHEMATICS 


6.  To  measure  the  distance  AB  across  a  swamp  (Fig.  129),  a 
man  walks  in  the  direction  AD,  so  that  /.BAD=  60°,  to  a  point 
C,  where  ^BCA  =  90°.    If  AC  =  300 yd., 

what  is  the  length  of  AB  ? 

7.  Find  the  number  of  degrees  in  each 
acute  angle  of  a  right  triangle  if  one  angle  is 

(a)  four  times  the  other ; 

(b)  three  fourths  of  the  other ; 

(c)  two  and  a  half  times  the  other ; 

(d)  5"  more  than  three  times  the  other ; 

(e)  5°  less  than  four  times  the  other. 

*8.  Practice  drawing  angles  of  30°,  45°,  60°,  and  90°  by 
using  wooden  or  cardboard  triangles. 

Ex.  5  illustrates  the  truth  of  the  theorem  In  a  right 
triangle  whose  acute  angles  are  30°  and  60°  the  side  opposite 
the  30°  angle  is  one  half  the  hypotenuse.  This  theorem  will 
be  proved  formally  later.  It  is  very  important  because  of 
its  many  practical  applications  in  construction  work  and 
elsewhere. 

189.  Isosceles  triangle ;  base  angles.  A  triangle  which 
has  two  equal  sides  is  called  an  isosceles  triangle.  The 
angles  opposite  the  equal  sides  are  called  the  base  angles  of 
the  isosceles  triangle. 

EXERCISES 

1.  Two  equal  acute  angles  of  a  right  triangle  are  repre- 
sented by  2  x  +  5  and  3  x  —  15.    Find  the  size  of  each  angle. 

2.  Draw  a  right  triangle  ABC 
(Fig.  130).    Draw  a  line  from 
C±AB;  call  the  foot  of  the  per- 
pendicular P.    Show  that  the 
perpendicular  (CP)  divides  the 

A  ABC  into  two  right  triangles.  FIG.  130 


EQUATION  APPLIED  TO  THE  TRIANGLE     139 

3.  In  Ex.  2  the  angle  x  is  the  complement  of  what  two 
angles  ?    What  is  the  relation  between  these  two  angles  ? 

\4\  In  Ex.  2,  /.y  is  the  complement  of  two  angles.    Indicate 
them.    Show  that  /. 


5.  Draw  freehand  three  isosceles  triangles. 

190.  Scalene  triangle.    A  scalene  triangle  is  a  triangle 
no  two  of  whose  sides  are  equal. 

EXERCISES 

1.  Draw  freehand  a  scalene  triangle. 

2.  Do  you  think  that  a  right  triangle  whose  acute  angles 
are  30°  and  60°  is  a  scalene  triangle  ?    Support  your  answer. 

3.  Draw    an     obtuse- 
angled   scalene  triangle. 

191.  Exterior  angles 
of    a    triangle.    If    the 
three    sides    of    a    tri- 
angle are  extended,  one 

at   each    vertex,   as    in    '' 

Fig.  131,  the  angles  thus     FlG-  13L   I^STRATING  THE  EXTERIOR 

ANGLES  OF  A  TRIANGLE 
formed    (x,  y,    and    z) 

are  called  exterior  angles  of  the  triangle  ABC. 

EXERCISES 

1.  How  many  exterior  angles  can  be  drawn  at  each  vertex 
of  a  triangle  ? 

2.  How  many  interior  angles  has  a  triangle  ?    How  many 
exterior  angles  ? 

3.  Draw  a  triangle  and  extend  the  sides  as  in  Fig.  131. 
Measure  the  three  exterior  angles  with  a  protractor.    What  is 
their  sum  ? 


14<» 


GENERAL  MATHEMATICS 


FIG.  132 


4.  I>ra\\  another  triangle  and  extend  the  sides  as  in.  Fig.  131. 
Cut  out  the  exterior  angles  (taking  one  at  each  vertex)  with  a 
pair  of  scissors  and  place  them  next  to 

each  other  with  their  vert  ices  together. 
What  does  their  sum  seem  to  be'/ 

5.  Find   the   sum   of  the  three   ex- 
terior angles  x,  //.  and  .-.-  in  Fig.  132  In- 
rotating  a  pencil   as    indicated   by  the 

arrowheads. 

6.  Show  that  tin-  mini  nf  flic  e;rti'ri<n- 

niKjli-x  i  >f  a  triangle  (taking  one  at  each  vertex)  /\  -><>0°  (two 
straight  angles). 

HINT.    How  many  degrees  are  in  the  sum  x  +  w/?    //  +  n  ?  z  +  r? 
(S,-,.  Fig.  l:;i.) 

Show  that  tlie  sum 

(.'•  +  m)  +  (//  +  n)  +  (z  +  ?•)  =  3  X  180°  =  540°. 
Then  this  fact  may  be  expressed  as  follows  : 

(x  +  ?/  +  r)  +  (HI  +  n  +  r)  =  540°. 

But  (x  +  y  +  z-)=  180°.  ^  Why  ? 

Therefore  (m  +  n  +  r)  =  :5i;o  .  \Vliy? 

7.  The  three  interior  angles  of  a  triangle  are  equal.    Find 
the  size  of  each  interior  and  each  exterior  angle. 


FIG.  133 


8.  Find  the  value  of  the  interior  and  exterior  angles  in  the 
triangle  of  Fig.  133. 


EQUATION  APPLIED  TO  THE  TRIANGLE     141 


FICJ.  134 


9.  Show  that  the  exterior  angle  x  of  the  triangle  AB(.' 
in  Fig.  134  is  equal  to  the  sum  of  the  two  nonadjacent  interior 
angles  A  and  C. 

10.  Using  Fig.  135,  in  which 
BD  II  ,1  (7,    prove    that   an    ex- 
terior   (inrjle    of  a    triangle    is      /  -\  \x 
('ijiinl   f'>   flif    sifin    of  the   two  A 

nonadjficent      interior      angles. 

Note  that  two  different  methods  are  suggested  by  this  figure. 

11.  Prove  Ex.  10  by  drawing  a  line  through  C  parallel  to  AB. 
HINT.    Extend  line  A  C. 

12.  Draw  a  quadrilateral.    Tear  off  the  corners  and  place 
the    interior   angles 

next  to  each  other 
by  the  method  of 
Ex.  2,  Art.  180. 
What  does  the  sum 
of  the  interior  an- 
gles seem  to  be  ? 

13.  Draw  a  quadrilateral  as  in  Fig.  136.    Draw  the  diagonal 
.4  C.  This  divides  the  quadrilateral  into  two  triangles.   What  is 
the  sum  of  the  interior  angles 

in  each  triangle  ?  What,  then, 
is  the  sum  of  the  interior 
angles  of  a  quadrilateral  ? 

14.  Draw  a  quadrilateral  as 
in  Fig.  136.   Produce  each  side 
(one   at  each   vertex).    What 
do  you  think  is   the  sum  of 
the    exterior    angles    of    the 

quadrilateral  ?    Check  your  estimate  by  measuring  the  angles. 

15.  Find  the  angles  of  a  quadrilateral  in  which  each  angle 
is  25°  smaller  than  the  consecutive  angle. 


FIG.  136 


142  GENEEAL  MATHEMATICS 

16.  Prove  that  the  consecutive  angles  of  a  parallelogram 
are  supplementary ;  that  is,  prove  x  +  y  =  180°,  in  Fig.  137. 

'17.  Prove    that    the 

opposite  angles  of  a  par-         _Dl /<?„___» 

allelogram  are  equal. 

HINT.    In  Fig.  137 show 
that  Zx  =  Za  =  Zz. 

18.  If   one   angle   of 

a  parallelogram  is  twice         »'  FIG.  137 

as  large  as  a  consecutive 

angle,  what  is  the  size  of  each  angle  in  the  parallelogram? 

19.  The  difference  between  two   consecutive  angles  of  a 
parallelogram  is  30°.    Find  the  size  of  all  four  angles  in  the 
parallelogram. 

20.  Show  that  the  sum  of  the  interior  angles  of  a  trapezoid 
is  two  straight  angles  (180°). 

21.  Prove  that  two  pairs  of 
consecutive  angles  of  a  trape- 
zoid are  supplementary.    (Use 
Fig.  138.) 

22.  In  Fig.  138,  Z  D  is  40°  Fl0'  138 

more  than  Z.  A,  and  /.B  is  96°  less  than  Z.  C.   Find  the  number 
of  degrees  in  each  angle. 

192.  The  construction  of  triangles.  We  shall  now  proceed 
to  study  three  constructions  which  require  the  putting 
together  of  angles  and  line  segments  into  some  required 
combination..  With  a  little  practice  the  student  will  see 
that  the  processes  are  even  simpler  than  the  thinking 
involved  in  certain  games  for  children  which  require  the 
various  combinations  of  geometric  forms. 

These  constructions  are  very  important  in  all  kinds  of 
construction  work ;  for  example,  in  shop  work,  mechanical 


EQUATION  APPLIED  TO  THE  TRIANGLE     143 

drawing,  engineering,  and  surveying.    The  student  should 
therefore  master  them. 

193.  Construction  problem.   To  construct  a  triangle  when 
the  three  sides  are  given. 

Construction.  Let  the  given  sides  be  a,  b,  and  c,  as  shown  in  Fig.  139. 

Draw  a  working  line  X  Y,  and  lay  off  side  c,  lettering  it  AB.  With 

A  as  a  center  and  witk  a  radius  equal  to  b  construct  an  arc  as  shown. 


c 
FIG.  139.    How  TO  CONSTRUCT  A  TRIANGLE  WHEN  THREE  SIDES 

ARE    GIVEN 

With  B  as  a  center  and  with  a  radius  equal  to  a  construct  an  arc 
intersecting  the  first.  Call  the  point  of  intersection  C.  Then  the 
triangle  is  constructed  as  required. 

EXERCISES 

v  1.  Construct  triangles  with  the  following  sides  : 
-  (a)  a  =  5  cm.,     b  =  5  cm.,     c  =  8  cm. 

(b)  a  =  1  cm.,     b  =  8  cm.,     c  =  4  cm. 

(c)  a  =  7  cm.,     b  =  9  cm.,     c  =  3  cm. 

2.  Is  it  always  possible  to  construct  a  triangle  when  three 
sides  are  given  ?  a 

3.  Construct    a     triangle, 

using    the     sides     given    in         | 1 

Fig.  140. 

. c 

4.  Compare  as  to  size  and 

,  ,  .       ,     ,  FIG.  140 

shape  the  triangle  drawn  by 

you  for  Ex.  3  with  those  drawn  by  other  pupils.  (See  if  the 
triangles  will  fold  over  each  other.) 


1  (JKNKKAL   MATHEMATICS 

5.  Make  a  wn.id.-n  triangle  by  nailing  three  sticks  together. 
Is  it  possible  to  change  the  shape  of  the  triangle  without  break- 
ing a  stick  or  removing  the  corner  nails  '/ 

6.  A  great  deal  of  practical  use  is  made  of  the  fact  that 
a  triangle  is  a  rigid  figure ;  for  example,  a  rectangular  wooden 
gate  is  usually  divided  into  two  triangles  by  means  of  a  wooden 
diagonal  so  as  to  make  the  gate  more  stable  (less  apt  to  sag). 
Try  to  give  other  examples  of  the  practical  use  that  is  made 

^of  the  stability  of  the  triangular  figure. 

7.  Construct  an  isosceles  triangle  having  given  the  base 
and  one  of  the  two  equal  sides. 

HINT.    Use  c  in  Fig.  140  as  the  base  and  use  //  twice:  that  is,  in 
Uais  case  take  a  =  />. 

8.  Measure  the  base  angles  of  the  isosceles  triangle  drawn 
for  Ex.  7.    What  appears  to  be  the  relation  between  the  base 

.angles  of  an  isosceles  triangle  '.' 

9.  Make  tracings  of  the  base  angles  drawn  for  Ex  7  and 
attempt  to  fold  one  angle  over  the  other.    Do  the  two  angles 
appear  to  represent  the  same  amount  of  rotation  ? 

10.  Compare   your   results   witli    those    obtained    by  your 
classmates. 

NOTE.    Results  obtained  from  Exs.  7-10  support  the  following 
theorem  :    Tin-  buxe  anglt-x  of  an  isosceles  triangle  fire  ef/>inl. 

11.  Construct  an  equilateral  triangle  having  given  a  side. 

12.  Study  the  angles  of  an  equilateral  triangle  by  pairs  in 
the  manner  suggested  by  Exs.  8-9.    State 

the  theorem  discovered. 

13.  To    measure     the     distance    AB 
(Fig.  141)  we  walk   from   B  toward  M 

so  that  Z/J  =  50°,  until  we  reach  C,  a  / 

point  at  which  /.ACB  =  50°.   What  line 

must  we  measure  to  obtain  A  B  ?   Why  ?  FIG.  141 


EQUATION  APPLIED  TO  THE  T1UANULE     145 


14.  At  a  point  C  (Fig.  142),  70  ft.  from  the  foot  of  a 
pole  AB,  the  angle  ACB  was  found  to  be  45°.  How  high  is 
the  pole  ? 

*15.  Walking  along  the  bank  of  a  river  from  A  to  C 
(Fig.  143),  a  surveyor  measures  at  B  the  angle  ABD,  and 
walking  500  ft.  further  to  C,  he  finds  angle  BCD  to  be  one 
half  of  angle  A  BD.  What  is  the  distance  BD  ? 

D 


C  A 

FIG.  142 


FIG.  143 


194.  Construction  problem.  To  construct  a  triangle,  hav- 
ing given  two  sides  and  the  angle  included  between  the 
sides. 

Construction.  Let  the  given  sides  be  a  and  b  and  the  given  angle 
be  Z  C,  as  shown  in  Fig.  144. 


B 


FIG.  144.    How  TO  CONSTRUCT  A  TRIANGLE  WHEN  Two  SIDES  AND  THE 
INCLUDED  ANGLE  ARE  GIVEN 

Draw  a  working  line  XY,  and  lay  off  CA  equal  to  b.  At  C  draw 
an  angle  equal  to  angle  C  by  the  method  of  §  78.  With  C  as  a 
center  and  a  radius  equal  to  a  lay  off  CB  equal  to  a.  Join  B  and  A, 
and  the  triangle  is  constructed  as  required. 


146 


GENERAL  MATHEMATICS 


EXERCISES 

1.  Is  the  construction  given  in  Art.  194  always  possible  ? 

2.  Draw  triangles  with  the  following  parts  given  :' 

(a)  a  =  3  cm.,     b  =  4  cm.,     Z  C  =  47°. 

(b)  c  =  lin.,      6  =  2i 

(c)  J  =  ljin.,    c  =  lf 

3.  Construct  a  triangle  with  the  parts  given  in  Fig.  145. 


Z.I  =  112°. 

=  87°. 


FIG.  145 

4.  Compare  as  to  size  and  shape  the  triangle  drawn  for  Ex.  3 
with  those  drawn  by  other  students  in  your  class.  (Place  one 
triangle  over  the  other  and  see  if  they  fit.) 

195.  Construction  problem.  To  construct  a  triangle  when 
two  angles  and  the  side  included  between  them  are  given. 


FIG.  146.   How  TO  CONSTRUCT  A  TRIANGLE  WHEN  Two  ANGLES  AND  THE 

SlDE    INCLUDED    BETWEEN    THEM    ARE    GIVEN 


Construction.    Let  ZA  and  ZB  be  the  given  angles  and  line  c  be 
the  given  included  side  (Fig.  146). 


EQUATION  APPLIED  TO  THE  TRIANGLE     147 

Lay  down  a  working  line  A'Fand  lay  off  AB  equal  to  linec  on^it. 
At  A  constrict  an  angle  equal  to  the  given  angled;  at  B  construct 
an  angle  equal  to  the  given  angle  B  and  produce  the  sides  of  those 
angles  till  they  meet  at  C,  as  shown.  Then  the  t\ABC  is  the 
required  triangle. 

EXERCISES 

1.  Draw  triangles  with  the  following  parts  given  : 
(a)  Z  A  =  30°,       Z  B  =  80°,     e  =  4  cm. 


(b)  Z  C  =  110°,     Z£  =  20°,     a  =  2  in. 

2.  Draw  a  triangle  with  the  parts  as  given  in  Fig.  147. 

3.  Is  the  construction  of  Ex.  2  always  possible  ? 


FIG.  147 

4.  Compare  as  to  size  and  shape  the  triangle  drawn  for  Ex.  2 
with  those  drawn  by  other  members  of  your  class.  (Fold  them 
over  each  other  and  see  if  they  fit.) 

SUMMARY 

196.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  right  triangle,  cardboard  triangle, 
wooden  triangle,  set  square,  isosceles  triangle,  scalene 
triangle,  interior  angles  of  a  triangle,  exterior  angles  of 
a  triangle,  base  angles. 

The  following  notations  have  been  given :  notation 
for  the  angles  and  sides  of  triangles,  notation  for  right 
triangle  (rt.  A). 


148 

197.  This  chapter  has  presented  methods  of 

1.  Finding  the  sum  of  the  interior  angles  of  a  triangle. 
±  Finding  the  sum  of  the  exterior  angles  of  a  triangle. 

3.  Drawing  right  triangles  by  means  of  wooden  or  card- 
board triangles. 

4.  Drawing  parallel  lines  by  means  of  the  wooden  tri- 
angle or  the  set  square. 

198.  This  chapter  has  taught  the  pupil  the  following 
constructions : 

1.  Given  three  sides  of  a  triangle,  to  construct  the  triangle. 

2.  Griven  two  sides  and  the  included  angle  of  a  triangle, 
to  construct  the  triangle. 

3.  Given  two  angles  and  the  included  side  of  a  triangle, 
to  construct  the  triangle. 

199.  The  following   theorems   have   been  presented   in 
this  chapter: 

1.  The  sum  of  the  interior  angles  of  a  triangle  is  a 
straight  angle  (180°). 

2.  The  sum  of  the  exterior  angles  of  a  triangle  is  two 
straight  angles  (360°). 

3.  If  two  angles  of  one  triangle  are  equal  respectively 
to  two  angles  of  another  triangle,  the  third  angle  of  the 
first  triangle  is  equal  to  the  third  angle  of  the  second. 

4.  The  acute  angles   of  a   right  triangle    are    comple- 
mentary. 

5.  In  a  right  triangle  whose  acute  angles  are  30°  and 
60°  the  side  opposite  the  30-degree  angle  is  one  half  the 
hypotenuse. 

6.  An  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  two  nonadjacent  interior  angles. 

7.  The  sum  of  the  interior  angles  of  a  quadrilateral  is 
four  right  angles  (360°). 


8.  The  sum  of  the  exterior  angles  of  a  quadrilateral 
is  four  right  angles  (360°). 

9.  The  opposite  angles  of  a  parallelogram  are  equal. 

10.  The    consecutive    angles    of    a   parallelogram    are 
supplementary. 

11.  Two  pairs  of  consecutive  angles  of  a  trapezoid  are 
supplementary . 

12.  The  base  angles  of  an  isosceles  triangle  are  equal. 

13.  An  equilateral  triangle  is  equiangular  (all  angles 
equal). 


CHAPTER  VIII 

POSITIVE  AND  NEGATIVE  NUMBERS.    ADDITION  AND 
SUBTRACTION 

200.  Clock  game.  Mary  and  Edith  were  playing  with  a 
toy  clock.  Each  took  her  turn  at  spinning  the  hand.  The 
object  of  the  game  was  to  guess  the  number  on  which  the 
hand  of  the  clock  stopped.  A  correct  guess  counted  five 
points.  If  a  player  missed  a  guess  by  more  than  three, 
she  lost  three  points.  If  she  came  within  three  she  either 
won  or  lo'st  the  number  of  points  missed,  according  to 
whether  she  had  guessed  under  or  over  the  correct  number. 
After  five  guesses  they  had  the  following  scores  : 

MARY  Solution.    The    score   as   kept   by  the 

Won     2  players  (the  words  are  inserted)  appeared 

Lost      1  as  follows : 

Lost  3 
Won  2 
Won  1 

EDITH 
Lost  1 
Lost  2 
Won  5 
Won  2 
Lost  2 

Who  won  the 

game?  Edith  won  the  game,  2  to  1. 

150 


MARY 

EDITH 

First  score 

0 

First  score 

CD 

Lost 

1 

Lost 

2 

Second  score 

1 

Second  score 

(D 

Lost 

3 

Won 

5 

Third  score 

d) 

Third  score 

2 

Won 

2 

Won 

2 

Fourth  score 

0 

Fourth  score 

4 

Won 

1 

Lost 

2 

Final  score 

1 

Final  score 

2 

POSITIVE  AND  NEGATIVE  NUMBERS        151 

201.  Positive  and  negative  numbers.    Algebraic  numbers. 
The  adding  of  scores  in  many  familiar  games  like  the  one 
cited  above  illustrates  an  extension  of  our  idea  of  count- 
ing that  will  be  found  very  useful  in  our  further  study  of 
mathematics.    It  is  important  to  notice  that  the  players 
began  at  zero  and  counted  their  scores  in  both  directions. 
Thus,  Mary  began  with  zero  and  won  two  points,  and  the 
2  she  wrote  meant  to  her  that  her  score  was  two  above 
zero.    On  the  other  hand,  Edith  had  zero  and  lost  one. 
In  order  to  remember  that  she  was  "  one  in  the  hole " 
she  wrote  1  within  a  circle.    On  the  next  turn  she  lost 
two  more,  and  she  continued  to  count  away  from  zero 
by  writing  3  within  a  circle.    The  same  idea  is  shown  in 
Mary's  score.    Her  second  score  was  1.    On  the  next  turn 
she  lost  three.    She  subtracted  3  from  1.    In  doing  so  she 
counted  backward  over  the  zero  point  to  two  less  than  zero. 
In  writing  the  scores  it  was  necessary  to  indicate  whether 
the  number  was  above  or  below  zero. 

We  shall  presently  have  numerous  problems  which 
involve  pairs  of  numbers  which  possess  opposite  qualities, 
like  those  above.  It  is  generally  agreed  to  call  num- 
bers greater  than  zero  positive  and  those  less  than  zero 
negative.  Such  numbers  are  called  algebraic  numbers. 

The  opposite  qualities  involved  are  designated  by  the 
words  "  positive  "  and  "  negative."  In  the  preceding  game, 
numbers  above  zero  are  positive,  whereas  numbers  below 
zero  are  negative. 

202.  Use  of  signs.    To  designate  whether  a  number  is 
positive  or  negative  we  use  the  plus  or  the  minus  sign. 
Thus,  4-  4  means  a  positive  4  and  —  4  means  a  negative  4. 
The  positive  sign  is  not  always  written.    When  no  sign  pre- 
cedes a  number  the  number  is  understood  to  be  a  positive 
number.    Thus,  3  means  +  3. 


152 

The  following  stock  quotations  from  the  Chicago  Daily 
Tribune  (March  24,  1917)  illustrate  a  use  of  the  plus  and 

minus  signs : 

CHICAGO  STOCK  EXCHANGE 


SALES 

HIGH 

Low 

CLOSE 

NET 

American  Radiator     .    .    . 

20 

295J 

295 

296 

-2 

Commonwealth  Edison  .    . 

79 

136i 

136  J 

136  J: 

-1 

Diamond  Match  

40 

81 

81 

81 



Swift  &  Co  

515 

114^ 

142f 

1441 

+  ~'- 

Peoples  Gas     

397 

95" 

90^ 

91 

-  5 

Prest-o-Lite     

40 

1292 

129 

129 

-3 

Sears  Roebuck    

325 

192 

190* 

19H 

~  <> 

Booth  Fisheries  

20 

79 

79 

79 

+    T 

The  last  column  shows  the  net  gain  or  loss  during  the 
day;  for  example,  American  Radiator  stock  closed  two 
points  lower  than  on  the  preceding  day,  Swift  &  Co. 
gained  2|,  Peoples  Gas  lost  5,  etc.  The  man  familiar 
with  stock  markets  glances  at  the  first  column  to  see 
the  extent  of  the  sales  and  at  the  last  column  to  see 
the  specific  gain  or  loss.  To  check  the  last  column  one 
would  need  the  quotations  for  the  preceding  day. 

In  order  that  we  may  see  something  of  the  importance 
of  the  extension  of  our  number  system  by  the  preceding 
definitions,  familiar  examples  of  positive  and  negative 
numbers  will  be  discussed. 

203.  Geometric  representation  of  positive  numbers.  Origin. 
We  have  learned  in  measuring  a  line  segment  that  when  a 
unit  (say^[_^)  is  contained  five  times 'in  another  segment, 
the  latter  is  five  units  long.  (In  general,  if  it  is  contained 
a  times,  the  measured  segment  is  a  units  long.)  We  may 
also  say  that  the  two  segments  represent  the  numbers  1  and 
o  respectively.  This  suggests  the  following  representation 


POSITIVE  AND  NEGATIVE  NUMBERS        153 

of  positive  integers :  On  any  line  OX  (Fig.  148)  beginning 
at  0  (called  the  origin  of  the  scale)  but  unlimited  in  the 
direction  toward  A',  we  lay  off  line  segments  OA,  AB,  BC, 
CD,  DE,  etc.,  each  one  unit  in  length.  We  thus  obtain  a 
series  of  points  on  the  o  A  B  c  D  E  F 

line,   each   of  which     I 1 1 \ 1 1 1 1 1 x 

corresponds  to  some    012345678 
positive   integer,  and     FlG-  148-   GEOMETRIC  REPRESENTATION  OF 
T,r        ,  POSITIVE  NUMBERS 

vice  versa.     We  also 

note  that  the  line  segment  which  connects  the  origin  with 
some  point  (say  JE")  is  the  corresponding  number  of  units 
in  length.  (Thus,  OE  =  5.)  Hence  the  positive  integers 
+  1,  -f  2,  -f  3,  +  4,  4-  5,  etc.  are  represented  by  OA,  OB, 
OC,  OD,  OE,  etc. 

204.  Geometric  representation  of  negative  numbers.  Num- 
ber scale.  If  we  prolong  OX  to  the  left  from  0  in  the 
direction  OX'  (Fig.  149),  we  may  lay  off  line  segments 
in  either  of  two  opposite  directions.  Segments  OC  and 
OD'  differ  not  only  hi  length  but  also  in  direction.  This 
difference  in  direction  is  indicated  by  the  use  of  the 

D'      C'     B1     A'     0       A       B       C       D 

Xr 1 1 1 1 1 1 1 1 1- 

-4-3-2-1       0        1       2       3       4 

FIG.  149.    GEOMETRIC  REPRESENTATION  OF  NEGATIVE  NUMBERS 

signs  +  and  — .  Thus,  OC  represents  +  3,  whereas  OD'  rep- 
resents —  4.  The  integers  of  algebra  may  now  be  arranged 
in  a  series  on  this  line  called  the  number  scale,  beginning 
at  0  and  extending  both  to  the  right  and  to  the  left. 
In  order  to  locate  the  points  of  the  number  scale  (Fig.  149) 
we  need  not  only  an  integer  of  arithmetic  to  determine 
how  far  the  given  point  is  from  zero  but  also  a  sign  of 
quality  to  indicate  on  which  side  of  0  it  is  found. 


154  GENEKAL  MATHEMATICS 

EXERCISES 

1.  Locate  the  following  points  on  the  number  scale  :  +  3, 
-3,  -7,  +  12,  -8,  +6. 

2.  If  we  imagine  that  each  of  the  players  in  the  clock 
game  (Art.  200)  has  a  string  or  tape  measure  graduated  to 
the  number  scale,  with  a  ring  on  it,  they  could  add  their 
scores  from  time  to  time  by  sliding  the  ring  back  and  forth. 

Indicate  on  the  number  scale  how  the  sum  of  the  following 
consecutive  scores  could  be  found  :  Began  at  0,  lost  1,  lost  2, 
won  5,  won  2,  lost  1. 

3.  Starting  from  the  middle  of  the  field  in  a  certain  football 
game,  the  ball  shifted  its  position  in  yards  during  the  first 
fifteen  minutes  of  play  as  follows:  .+  45,  —15,  +11,  —10, 
—  2,  +  23.    Where  was  the  ball  at  the  end  of  the  last  play  ? 

205.  Addition  and  further  use  of  positive  and  negative 
numbers.  The  preceding  exercises  show  that  positive  and 
negative  numbers  may  be  added  by  counting,  the  direc- 
tion (forward  or  backward)  in  which  we  count  being 
determined  by  the  sign  (+  or  —  )  of  the  numbers  which 
we  are  adding.  Thus, 

To  add  +  4  to  +  5  on  the  number  scale  of  Fig.  149,  begin  at  +  5 
and  count  4  to  the  right. 

To  add  —  4  to  +  5  begin  at  +  5  and  count  4  to  the  left. 

To  add  —  4  to  —  5  begin  at  —  5  and  count  4  to  the  left. 

To  add  +  4  to  —  5  begin  at  —  5  and  count  4  to  the  right. 

The  results  are  as  follows  : 

+  5  +(+4)  =  +  9;     +5+(-4)  =  +  l;       -5  +  (-4)  =  -9; 


+  5  +  (—  4)  =  -f  1  is  read  "positive  5  plus  negative  4  equals 
positive  1." 

—  5  +(—4)  =  —9  is  read  "negative  5  plus  negative  4  equals 
negative  9." 


•  •7 


POSITIVE  AND  NEGATIVE  NUMBERS        155 

EXERCISES 

1.  Give  the  sum  in  each  of  the  following.    Be  prepared  to 
interpret  the  result  on  the  number  scale. 

-(a)  3 +(+2).  (g)  6+(-l).  -  (m)  -2+(-5). 

(b)  4  +(-  3).  (h)  6  +(-  3).  ^(n)  -  4  +  6. 

2.  On  a  horizontal  straight  line,  as  X'OX  in  Fig.  150,  con- 
sider the  part  OX  as  positive,  and  the  part  OX'  as  negative. 
Construct  line  segments  correspond- 
ing to  the  following  numbers  :  —  2, 

+  6,  -  3,  -  5,  0. 

3.  Consider  OF  as  positive  and  OY'     xr 
as  negative  and  construct  segments  on 

the  line  YOY'  corresponding  to  —  4, 
-  2,  +  3,  +  4,  -  3,  0. 

4.  A  bicyclist  starts  from  a  certain 

point  and  rides  18  mi.  due  northward  -pIG  JCQ 

(4- 18  mi.),    then    12  mi.    due    south- 
ward (— 12  mi.).    How  far  is  he   from  the  starting  point? 

5.  How  far  and  in  what  direction  from  the  starting  point  is 
a  traveler  after  going  eastward  (+)  or  westward  (— )  as  shown 
by  these  pairs  of  numbers :  +  16  mi.,  then  —  3  mi.?  —  2  mi.,  then 
-f-  27  mi.?  —  16  mi.,  then  +  16  mi.?  +  100  mi.,  then  4-  52  mi.? 

6.  Denoting  latitude  north  of  the  equator  by  the  plus  sign 
and  latitude  south  by  the  minus  sign,  give  the  meaning  of  the 
following  latitudes  :  4-  28°,  +  12°,  -  18°,  +  22°,  -  11°. 

7.  Would  it  be  definite  to  say  that  longitude  east  of  Green- 
wich is  positive  and  west  is  negative  ?    What  is  the  meaning 
of  the  following  longitudes':  4-  42°?  +  142°?  -  75°?  -  3°  ? 


156  GENERAL  MATHEMATICS 

8.  A  vessel  starting  in  latitude  +  20°  sails  +  17°,  then  —  G3°, 
then  +  42°,  then  —  1G°.    What  is   its   latitude  after  all  the 
sailings  ? 

9.  What)  is  the  latitude  of  a  ship  starting  in  latitude  —  53° 
after  the  following  changes  of  latitude :  +  12°,  - 15°,  +  28°, 
-  7°,  +  18°,  -  22°,  +  61°  ? 

206.  Double  meaning  of  plus  and  minus  signs.    Absolute 
value.    The  many  illustrations  of  negative  number  show 
that  there  is  a  real  need  in  the  actual  conditions  of  life 
for  the  extension  of  our  number  scale  so  as  to  include 
this  kind  of  number.    It  must  be  clear-ly  understood  that 
a  minus  sign  may  now  mean  two  entirely  different  things : 
(1)  It  may  mean  the  process  of  subtraction,  or  (2)  it  may 
mean  that  the  number  is  a  negative  number.    In  the  latter 
case  it  denotes  quality.    Show  that  a  similar  statement  is 
true  for  the  plus  sign. 

This  double  meaning  does  not  often  confuse  us,  since  it  is 
usually  possible  to  decide  from  the  context  of  the  sentence 
which  of  these  two  meanings  is  intended.  Sometimes  a 
parenthesis  is  used  to  help  make  the  meaning  clear,  thus 
4°  +  (—  3°)  means  add  a  negative  3°  to  a  positive  4°. 

Sometimes  we  wish  to  focus  attention  merely  on  the 
number  of  units  in  a  member  regardless  of  sign.  In  that 
case  we  speak  of  the  absolute  value,  or  numerical  value. 
Thus,  the  absolute  value  of  either  -f  4  or  —  4  is  4. 

207.  Forces.    In  mechanics  we  speak   of  forces  acting 
in  opposite  directions  as  positive  and  negative.     Thus,  a 
force  acting  upward  is  positive,  one  acting  downward  is 
negative. 


POSITIVE  AND  NEGATIVE  NUMBERS         157 


EXERCISES 

1.  Three  boys  are  pulling  a  load  on  a  sled,  one  with  a  force 
of  27  lb.,  another  with  a  force  of  56  lb.,  and  the  third  with  a 
force  of  90  lb.    With  what  force  is  the  load 
being  pulled  ? 

2.  Two  small  boys  are  pulling  a  small 
wagon  along;  one  pulls  with  a  force  of 
23  lb.,  and  the  other  pulls  with  a  force  of 
36  lb.  A  boy  comes  up  behind  and  pulls 
with  a  force  of  47  lb.  in  the  opposite  direc- 
tion from  the  others.  What  is  the  result  ? 


FIG.  151 


3.  An  aeroplane  that 
can  fly  48.3  rni.  an  hour 
in  still  air  is  flying 
against  a  wind  that 

retards  it  19.6  mi.  an  hour.    At  what  rate 

does  the  aeroplane  fly  ? 

4.  A  balloon  which  exerts  an  upward 
pull  of  512  lb.  has  a  453-pound  weight 
attached  to  it.    What  is  the  net  upward 
or  downward  pull  ? 

5.  A  toy  balloon  (Fig.  151)  tends  to  pull 
upward  with  a  force  of  8  oz.  What  happens 
if  we  tie  a  5-ounce  weight  to  it  ?  an  8-ounce  ? 

6.  A  boy  can  row  a  boat  at  the  rate  of 
4  mi.  per  hour.    How  fast  can  he  go  up 
a  river  flowing  at  the  rate  of  2^  mi.  per 
hour  ?    How  fast  can  he  ride  down  the 
river  ?    How  fast  could  he  go  up  a  stream 
flowing  5  mi.  per  hour  ? 

208.  The  thermometer.  The  ther- 
mometer (Fig.  152)  illustrates  the  idea 
of  positive  and  negative  numbers  in  two 


Z12- 


100- 


70- 


32- 


FIG.  152.  THE  TIIKK- 
SIOMETER  ILLUSTRATES 
THE  IDEA  OF  POSI- 
TIVE AND  NEGATIVE 

NUMBEBS 


158  GENERAL  MATHEMATICS 

ways.  In  the  first  place,  the  number  scale  is  actually  pro- 
duced through  the  zero,  and  degrees  of  temperature  are 
read  as  positive  (above  zero)  or  negative  (below  zero). 
In  the  second  place,  the  thermometer  illustrates  positive 
and  negative  motion  discussed  in  the  preceding  article. 
Thus,  when  the  mercury  column  rises,  its  change  may  be 
considered  as  positive,  +  5  indicating  in  this  case  not  a 
reading  on  the  thermometer,  as  before,  but  a  change  (rise) 
in  the  reading ;  similarly,  —  3°  in  this  sense  indicates  a 
drop  of  3°  in  the  temperature  from  the  previous  reading. 

EXERCISES 

1.  What   is   the   lowest  temperature  you  have  ever  seen 
recorded  ? 

2.  The  top  of  the  mercury  column  of  a  thermometer  stands 
at  0°  at  the  beginning  of  an  hour.    The  next  hour  it  rises  5°, 
and  the  next  3°;  what  does  the  thermometer  then  read  ? 

3.  If  the  mercury  stands  at  0°,  rises  8°,  and  then  falls  5°, 
what  does  the  thermometer  read  ? 

4.  Give  the  final  reading  in  each  case : 

A  first  reading  of  10°  followed  by  a  rise  of  2°. 
A  first  reading  of  10°  followed  by  a  fall  of  12°. 
A  first  reading  of  20°  followed  by  a  fall  of  18°. 
A  first  reading  of  x°  followed  by  a  rise  of  y°. 
A  first  reading  of  x°  followed  by  a  fall  of  y°. 
A  first  reading  of  a°  followed  by  a  rise  of  a°. 
A  first  reading  of  a°  followed  by  a  fall  of  a°. 
A  first  reading  of  —  a°  followed  by  a  fall  of  —  a°. 

5.  The  reading  at  6  P.M.  was  7°.    What  was  the  final  read- 
ing  if   the   following   numbers    express   the   hourly   rise   or 
fall :  +  2°,  + 1°,  0°,  -  3°,  -  3°,  -  2°,  -  2°,  -  1°,  -  1°,  -  3°, 
—  1°  4-  2°  ? 


POSITIVE  AND  NEGATIVE  NUMBERS        159 

6.  Add  the  following  changes  to  find  the  final  reading,  the 
first  reading  being  0° :  +  3°,  +  2°,  -  4°,  -  3°,  -  2°,  -f  3°. 

7.  The  differences  in  readings  of  a  thermometer  that  was 
read  hourly  from  5  A.  M.  until  5  P.  M.  were  as  follows :  3°,  4°, 
7°,  10°,  12°,  9°,  8°,  5°,  0°,   -22°,  -17°,  -12°.    How  did 
the  temperature   at   5  P.M.   compare  with 'that   at   5  A.M.? 
If  the. temperature  at  5  A.M.  was  -f-  20°,  make  a  table  showing 
the  temperature  at  each  hour  of  the  day. 

209.  Positive  and  negative  angles.  By  rotating  line  AB 
in  a  plane  around  A  until  it  takes  the  position  AC  the 
angle  BAG  is  formed  (Fig.  153). 
By  rotating  AB  in  the  opposite  direc- 
tion angle  BAC±  is  formed.  To  dis- 
tinguish between  these  directions 
one  angle  may  be  denoted  by  the 
plus  sign,  and  the  other  by  the  FlG  153 

minus  sign.    We  agree  to  consider 

an  angle  positive  when  it  is  formed  by  rotating  a  line 
counterclockwise  and  negative  when  it  is  formed  by  clock- 
wise rotation.  This  is  simply  another  illustration  of  motion 
in  opposite  directions. 

EXERCISES 

1.  In  this  exercise  the  sign  indicates  the  direction  of  rotation. 
Construct  the  following  angles  with  ruler  and  protractor,  start- 
ing with  the  initial  line  in  the  horizontal  position :  -f  30°,  -(-  45°, 
+  90°,  +43°,  +212°,  -30°,  -45°,  -90°,  -53°,  -182°, -36°. 

2.  Find  the  final  position  of  a  line  which,  starting  at  OX 
(horizontal),  swings  successively  through  the  following  rota- 
tions :  +  72°,  -  38°,  + 112°,  -  213°,  +  336°,  -  318°,  -  20°, 
+  228°. 

3.  Do  you  see  a  short  cut  in  finding  the  final  position  of 
the  line  in  Ex.  2  ? 


GENERAL  .MATHEMATICS 

210.  Business  relations.    Finally,   the   idea  of    positive 
and  negative  numbers  may  be  further  illustrated  by  the 
gain  or  loss  in  a  transaction ;  by  income  and  expenditure ; 
by  a  debit  and  a  credit  account ;  by  money  deposited  and 
money  checked  out;  and  by  the  assets  and  liabilities  of 
a  business  corporation.    Thus,  a  bankrupt  company  is  one 
which  has  not  been  able  to  prevent  the  negative  side  of  the 
ledger  from  running  up  beyond  the  limit  of  the  confidence 
of  its  supporters. 

EXERCISES 

1.  The  assets  of  a  company  are  $ 26,460,  and  its  liabilities 
are  $39,290.    What  is  its  financial  condition  ? 

2.  A  newsboy  having  $25  in  the  bank  deposits  $10.25  on 
Monday,  checks  out  $16.43  on  Tuesday,  checks  out  $7.12  on 
Wednesday,  deposits  $5  on  Thursday,  deposits  $7.25  on  Friday, 
and  checks  out  $11.29  on  Saturday.    What  is  his  balance  for 
the  week  ? 

3.  If  a  man's  personal  property  is  worth  $1100  and  his  real 
estate  $12,460,  and  if  his  debts  amount  to  $2765,  what  is  his 
financial  standing  ? 

4.  A  boy  buys  a  bicycle  for  $10.25  and  sells  it  for  $6.    Does 
he  gain  or  lose  and  how  much  ? 

211.  Addition  of  three  or  more  monomials.    The  following 
exercises  will  help  us  to  see  how  the  addition  of  monomials 
may  be  extended. 

EXERCISES 

1.  Add  the  following  monomials  : 

(a)  2  +  3  +  (-  4)  +  (5).          (c)  (-  4)  +  2  +  3  +  (-  5). 

(b)  3  +  2'+  (-  4)  +  (-  5).     (d)  (-  5)  +  (-  4)  +  3  +  2. 

2.  In  what  form  has  the  commutative  law  of  addition  been 
stated  ?    (Art.  36.)    Does  it  seem  to  hold  when  some  of  the 
addends  are  negative? 


POSITIVE  AND  NEGATIVE  NUMBERS        161 

3.  Show  by  a  geometric  construction  on  squared  paper  that 
the  commutative  law  holds  when  some  of  the  addends  are 
negative.  * 

HINT.  First  take  a  line  segment  —  a  units  long  (Fig.  154)  and 
add  a  segment  +  b  units  long  (Fig.  154)  ;  then  take  the  segment 
that  is  +  b  units  long  and  add  the  segment  Q 

that  is  —  a  units  long.    The  results  should      L^       ~a        I 
be  the  same. 


4.  Does  a  bookkeeper  balance  the  ac-     r~~          IfT^ 
count  every  time  an  entry  is  made  or  does  „  .   -.. 

he  keep  the  debits  and  credits  on  separate 
pages  and  balance  the  two  sums  at  the  end  of  the  month  '.' 

The  process  of  adding  several  positive  and  negative  num- 
bers can  be  explained  in  detail  by  the  following  problem  : 

Add  -  10,  +  50,  -  27,  +  18,  -  22,  -  31,  +  12. 

Arrange  the  numb^s  as  follows  :  —  10 

+  50 
-27 
+  18 
-22 
-31 
+  12 

There  are  three  positive  terms,  50,  18,  and  12,  whose  sum  is  80. 
There  are  four  negative  terms,  —  10,  —  27,  —  22,  and  —  31,  whose 
sum  is  -  90.  Adding  +  80  to  -  90  gives  -  10. 

The  process  consists  of  the  following  three  steps  : 

1.  Add  all  the  positive  terms. 

2.  Add  all  the  negative  terms. 

3.  Add  these  two  sums  by  the  following  process  :  (a)  De- 
termine how  much  larger  the  absolute  value  of  one  number  is 
than  the  absolute  value  of  the  other,     (b)    Write  this  number 
and  prefix  the  sign  of  the  greater  addend. 


ltli>  GENERAL  MATHEMATICS 

212.  Algebraic  addition.  The  results  of  the  preceding 
exercises  show  that  positive  and  negative  numbers  may 
be  added  according  to  the  f  ollo\f1ng  laws  : 

1.  To  add  two  algebraic  numbers  having  like  signs  find 
the  sum  of  their  absolute  'values  and  prefix  to  this  sum  their 
common  sign. 

2.  To  add  two  algebraic  numbers  having  unlike  signs  find 
the  difference  of  their  absolute  values  and  prefix  to  it  the  sign 
of  the  number  having  the  greater  absolute  value. 

EXERCISES 

1.  Show  that  the  sum  of  two  numbers  with  like  signs  is  the 
sum  of  their  absolute  values  with  the  common  sign  prefixed. 
Illustrate  with  a  concrete  experience. 

2.  Show  that  the  sum  of  two  numbers  having  unlike  signs 
but  the  same  absolute  value  is  zero.    Illustrate  with  some  fact 
from  actual  experience. 

3.  Find  the  following  sums,  performing  all  you  can  orally  : 

(a)  -5  (d)-7  (g)  -f  -(j)-17f* 
±1                 -«                     +j  +261* 

H 

(b)  +5  (e)  -Sa  (h)  -fai  (k)  +  62£z2 

-28f  ae8 


<<0  +  7 

(f) 

-ftjw 

(i) 

3.16  x 

(1) 

-2.3a;m2 

1  +5 

—  16m 

-5.28  a; 

+  6.5  xm* 

4-  3- 

Find  the  following  sums 

4-  --    6               5.  -f    3  6.  +51  7.   -242 

+  10                   +10  +23  +726 

•    8                      -7  -18  58 

4                   -    4  -7  +24 


POSITIVE  AND  NEGATIVE  NUMBERS         163 


8.  +    7.5 
+  12.5 
-    9.5 
+    2.5 

9.  --    Sx 
+    4z 
+  17  a; 


10.  +  7x 
-lOx 
-I2x 


12. 

+  t 

16.  +    7| 

20.  0.5  x2 

+  | 

—    7£ 

0.23  x* 

i 

—  2 

-i2i 

0.12  a;2 

-    2J 

0.07  x2 

13. 

T| 

17.   -  12.18  - 

21.  +27  a;2 

+  1 

-  11.88 

—  15  ar* 

_  3 

+  13.16 

+  17x2 

+  1 

-14.08- 

-12  a:2 

14. 

+  3| 

18.  +10.05 

22.  +23  am 

—  2  <s 

+    4.85 

—  14  xm 

—  5J 

-    3.25 

—  17  xm 

—  2I 

-  12.35 

+  20  asm 

15. 

+  4f 

19.  -3.1s 

23.  -  18.25  arm 

—  61 

-5.4s 

+  17.34  aftw 

-6f 

+  7.2s 

-  19.64  x2m 

4  8f 

-3.1s 

+  21.17  a;2m 

11.  +24  a 

—  6a 
-    7a 

—  3a 


213.  Drill  exercises.  The  following  exercises  constitute 
a  drill  in  determining  the  common  factor  of  similar  mono- 
mials and  applying  the  law  for  the  addition  of  similar 
monomials  (Art.  40).  We  need  to  recall  that  the  sum  of 
two  or  more  similar  monomials  is  a  number  whose  coefficient 
is  the  sum  of  the  coefficients  of  the  addends  and  whose  literal 
factor  is  the  same  as  the  similar  addends.  The  exercises 
are  the  same  as  a  preceding  set  (Art.  40)  except  that  in 
this  case  the  step  in  which  the  coefficients  of  the  addends 
are  added  involves  the  addition  of  positive  and  negative 
numbers. 


164  GENERAL  MATHEMATICS 

EXERCISES 

In  each  case  (1)  point  out  with  respect  to  what  factor  the 
following  terms  are  similar ;  (2)  express  as  a  monomial  by  add- 
ing like  terms : 

1.  3y,  -6y,  20y,  -  35y. 
Solution.    The  common  factor  is  y. 

The  sum  of  the  coefficients  is  3  +  (-  6)  +  20  +  (-  35)  =  -  18. 
Whence  the  required  sum  is  — 18  y. 

2.  5x,—7x,—9x,+12x,—3x. 

3.  7  b,  -  12  b,  -  9  b,  +  11  b,  -  13  b. 

4.  9  ab,  -  17  ab,  -  11  ab,  +  13  06. 

5.  —  8  mnx*,  +  12  wmx2,  —  15  wwia2,  —  13  mnx*. 

6.  5  a2*,  -  7  o%,  +  9  a2*,  -  5  ^b. 

7.  3  ax,  4  ox,  —  8  az,  —  7  or. 

8.  -  5/,r/,  lpq\  -  &spq*,  ~  Bpf. 

9.  az,  -  14  z,  -bz,  +12  z. 
Solution.    The  common  factor  is  :. 

The  sum  of  the  coefficients  is  a  +  (-  14)  +  (-&)  +  12. 
Since  a  and  —  fe  are  still  undetermined,  we  can  only  indicate  that 
sum  thus :  a  —  14  —  b  +  12. 

Whence  the  sum  written  as  a  monomial  is  (a  —  b  —  2)  z. 

10.  mx,  +  5  x,  —7x,  +  ex. 

11.  m/,  -  wf,  +  5  y2,  -  12  y8,  +  c/. 

12.  tyab,  +  4|aJ,  —  5^  ab,  -  6|  aft. 

214.  Addition  of  polynomials.  We  have  had  numerous 
examples  of  the  addition  of  polynomials  in  dealing  with  per- 
imeters. In  applying  the  principles  involved  to  polynomials 
having  positive  and  negative  terms  we  need  to  recall  that 
in  addition  the  terms  may  be  arranged  or  grouped  in  any 
order.  Thus, 

2  +  3  +  4  =  3  +  2  +  4  (Commutative  Law) 

5  +  (-  3)  +  4  =  -  3  +  (5  +  4)  (Associative  Law) 


POSITIVE  AND  NEGATIVE  NUMBERS         165 

In  adding  polynomials  it  is  convenient  to  group  similar 
terms  in  the  same  column,  much  as  we  do  in  adding 
denominate  numbers  in  arithmetic. 

EXERCISES 

1.  Add  the  following  polynomials  and  reduce  the  sum  to  its 
simplest  form  :   3  yd.  +  1  ft.  +  6  in.,  5  yd.  +  1  ft.  +  2  in.,  and 
12  yd.  +  3  in. 

Solution.   Writing  the  similar  terms  in  separate  columns  we  have 

3  yd.  1  ft.  6  in. 

5  yd.  1  ft.  2  in. 

12yd.  3  in. 

20yd.  2fE  11  in. 

Note  that  the  common  mathematical  factors  are  not  yards,  feet, 
inches  f  but  the  unit  common  to  all  of  them,  or  inches.  The  problem 
may  be  written  as  follows : 

3  x  36  +  1  x  12  +    6 

r>  x  36  +  1  x  12  +  2 
12  x  36  _  +  3 

20  x  36  +  2  x  12  + 11 

2.  Add  9  tj  +  3 x  +  2  /,  o  //  +  2 x  +  (>  I,  and  3  y  +  2  x  +  8  i. 

9  y  +  3  x  +  2  i 
5  y  +  2  x  +  6  ( 
3  y  +  2  x  +  81 


17  y  +  Tar  4-  16  i 

3.  Add    27  xs  —13xy+  1C  if',    -  14  a-8  +  25  xy  +  4  ?/,    and 


Solution.    Writing  similar  terms  in  separate  columns  and  adding, 

we  have 

27  Xs  -  13  xy  +  16  y2 

-  14  a-8  +  25  xy  +    4  ?/2 


-    6  xz/  +    8  y2 


166  GENERAL  MATHEMATICS 

In  the  following  exercises  add  the  polynomials  : 


+  2b  +  "c  7.  ~6x  +  15y-l6z-     Sw 

-9/;  -2c  -Sx  +  22y  +  l6z-12w 

+  3b  —  5c  3x+    -ly—    5z+    6w 

2b  —  3c  9x-     8+    7z—    6w 


5.        2x  +  3y  +  4«  8.   6r-2s  +  3t 

5x  —  6y—7z  5r            —3t 

4z  +  5y  +  8s  2r+6s-5t 

-2x-5y-2z  3s  +     t 


6.       2a  +  5b+7c  \9\       2x  —  3y  +  ±z 

—  3a  —  Sb  —  5c  x+    y  —  3z 


10.  —  2x  —  6b  +  13c,  —  llaj  +  19ft  —  30  c,  and  5x  +4c. 

11.  12A;-10Z  +  9m,  -27c  +  2^-4m,  and  - 

12.  14  e  —  3  y  +  1  z,  5  e  +  5  y  —  3  z,  and  6e  +  4:y  +  2z. 

13.  24z- 


NOTE.  Here  certain  terms  are  inclosed  in  grouping  symbols  (  ) 
called  parenthesis.  These  indicate  that  the  terms  within  are  to  be 
treated  as  one  number  or  one  quantity.  Other  grouping  symbols  will 
be  given  when  needed  (see  pp.  175,  177). 

14.  (6*- 

15.  3 


215.  Degree  of  a  number.  The  degree  of  a  number  is 
indicated  by  the  exponent  of  the  number.  Thus,  x2  is  of 
the  second  degree  ;  a-3,  of  the  third  degree  ;  y\  of  the  fourth 
degree  ;  etc.  The  monomial  3  xyh*  is  of  the  first  degree 
with  respect  to  x,  of  the  second  degree  with  respect  to  y, 
and  of  the  third  degree  with  respect  to  r. 


POSITIVE  AND  NEGATIVE  NUMBERS        167 

216.  Degree  of  a  monomial.    The  degree  of  a  monomial 
is  determined  by  the  sum  of  the  exponents  of  the  literal 
factors.    Another  way  of  saying  this  is :   The  number  of 
literal  factors  in  a  term  is  called  the  degree  of  the  term. 
Thus,  3?  is  of  the  second  degree ;  xy2,  of  the  third  degree ; 
and  4:ry222,  of  the  fifth  degree. 

EXERCISES 

Determine  the  degree  of  the  following  monomials : 

1.  2ajy.       3.  3 ft4.  5.  2±mxy.      7.  ^-       9.  rV. 

3.2  z 

2.  2abs.        4.  5x7/V.       6.  —  •  8.  rsl.      10.  m*x*ifz\ 

m 

217.  Degree  of  a  polynomial.  The  degree  of  a  polynomial 
is  determined  by  the  degree  of  the  term  having  the  highest 
degree.    Thus,  x^y*  +  x+%y  +  5  is  of  the  fourth  degree, 
and  5o^  —  a^  +  7isa  third-degree  expression. 

EXERCISES 

Indicate  the  degree  of  the  following  polynomials  : 

1.  x*  +  2x3  -  x4  +  2x*  +  y  +  7.          4.  x  -  2xif+  if. 

2.  x  +  2xij  +  if.     '  5.  a;4 +  4. 

3.  y?—1xy->ryi.  6.  x5  +  x3  +  x*  +  1. 

218.  Arrangement.    A  polynomial  is  said  to  be  arranged 
according  to  the  descending  powers  of  x  when  the  term  of 
the  highest  degree  in  x  is  placed  first,  the  term  of  next 
lower  degree  next,  etc.,  and  the  term  not  containing  x 
last.    Thus,  2  +  x-\-x8-\-^3^  when  arranged  according  to 
the  descending  powers  of  x  takes  the  form  z3+3z2  +  #+2. 
When  arranged  in  the  order  2  +  a;  +  3«2  +  «3,  the  polynomial 
is  said  to  be  arranged  according  to  the  ascending  powers  of  x. 


168  GENERAL  MATHEMATICS 

Find  the  sura  of  -3a2+2a3-4a,  -a2  +  7,  5a3-4a 
and  -2«3-7-2a2. 

Arranging  according  to  descending  powers  and  adding,  we  have 
2  a8  -  3  a2  -  4  a 

a2  +7 

5  a8  -  4  a  +  3 

-  2  a8  -  2  a2  -  7 

5a8-6a2-8a  +  3 

Check.  One  way  to  check  is  to  add  carefully  in  reverse  order,  as 
in  arithmetic. 

A  second  method  for  checking  is  shown  by  the  following : 

Let  a  =  2.   Then  we  have 

2  a3  —  3  a2  —  4  a          =  —    4 

a2  +  7  =        3 

5  a8  -  4  a  +  3  =      35 

-  2  a3  -  2  a2  -  7  =  -  31 

5a3-6a2-8a  +  3=        3 

The  example  checks,  for  we  obtained  3  by  substituting 
2  for  a  in  the  sum  and  also  by  adding  the  numbers  obtained 
by  substituting  2  for  a  separately  in  the  addends. 

EXERCISES 

In  the  following  list  arrange  the  polynomials  in  columns 
either  according  to  the  ascending  or  the  descending  order  of 
some  one  literal  factor.  Add  and  check  as  in  the  preceding 
problem. 

1.  x2  +  if  +  xy,  x2  -xy  +  y*,—  14 xy  +  3x*-2y'2. 

2.  26xi/,  -5y*  +  12x*,  -  IQxy  -  ISxy  +  16?f. 

3.  5.3  x2  -  13.6  xy  -  2.3  f,  -  0.02  f  +  5  xy  +  3.2  x2. 

4.  ±x8-Sx2-5x-12,  3x*  +  3x-5x*  +  8. 

5.  8  a3  -  2  a2  +  3  a  -  6,  -  3  «8  +  2  a2  +  a  +  7. 

6.  3  ?-2  +  2  r8  +  3  r  -  5,  -  r*  -  2  r  +  rs  +  1,  r2  -  2. 

7.  f  s2  -  i  ™  -  §  r2,  -  s*  _  §  rs  +  ^  s  f  -5r*. 


POSITIVE  AND  NEGATIVE  NUMBERS         169 

8.  -  21  oty  4-  40  a?»  +  55  *2?/  -  14  ^, 

+  2  cx2z  +  58  x«2  -  23  y8*, 


9.  4  (m  +  TO)  -  6  (m2  +  w2)  +  7  (m3  +  n3)  -  8(m4  +  n4), 

9  (ra4  +  w4)  -  3  (m8  +  w8)  +  4  O2  +  re3)  -  3  (m  +  »). 

10.  3  (a  +  i)  -  5  (a2  +  &)  +  7  (a  -  i2)  -  5  (a2  +  a&  +  62), 

6  (a  +  l>)  -  3  (a  -  62)  +  4  (a2  +  V)  +  2  (a2  +  ab  +  i2). 

219.  Subtraction.    The  following  exercises  will  help  to 
make  clear  the  process  of  subtraction. 

EXERCISES 

1.  If  the  thermometer  registers  +  24°  in  the  morning  and 
-f-  29°  in  the  evening,  how  much  warmer  is  it  in  the  evening 
than  in  the  morning  ?    How  do  you  find  the  result  ? 

2.  A  thermometer  registers  +  10°  one  hour  and  +  14°  the 
next  hour.    What  is  the  difference  between  these  readings  ? 

3.  If  the  thermometer  registers  —  2°  one  hour  and  +  3° 
the  following  hour,  how  much  greater  is  the   second  read- 
ing than  the  first  ?  The  question  might  be  stated  :    What  is 
the  difference  between  —  2°  and  +  3°  ?    In  what  other  form 
could  you  state  this  question? 

4.  If  you  were  born  in  1903,  how  old  were  you  in  1915? 
State  the  rule  you  use  in  finding  your  age. 

5.  The  year  of  Christ's  birth  has  been  chosen  as  zero  of 
the   time   scale   in   Christian   countries.    Thus,  we  record  a 
historic  event  as  60  B.C.  or  A.  D.  14.    Instead  of  using  B.C.  and 
A.D.  we  may  write  these  numbers  with  the  minus  and  plus 
signs  prefixed.    How  old  was  Caesar  when  he  died  if  he  was 
born  in  —  60  and  died  in  -f-  14  ? 

6.  A  boy  was  born  in  —  2.   How  old  is  he  in  +  5  ?    Apply 
your  rule  for  subtraction  in  this  case. 

7.  Subtract  —  3  from  +  5  ;  —  4  from  +  5  ;  —  5  from  +  5  ; 
-  50  from  4-  25. 


170  GENERAL  MATHEMATICS 

8.  A  newsboy  has  410.    How  much  must  he  earn  during 
the  day  so  as  to  have  850  in  the  evening?    State  the  rule 
which  you  use  to  solve  problems  of  this  kind. 

9.  A  newsboy  owes  three  other  newsboys  a  total  of  650. 
How  much  must  he  earn  to  pay  his  debts  and  have  200  left? 
Apply  your  rule  for  solving  Ex.  8  to  this  problem. 

10.  John  is  $25  in  debt,  Henry  has  #40  in  cash.   How  much 
better  off  is  Henry  than  John  ?  Apply  the  rule  stated  for  Ex.  8. 

11.  What  is  the  difference  between  12  and  20?  5  and  45  ? 
0  and  20  ?  -  1  and  20  ?  -  5  and  25  ?  -  12  and  18  ?      . 

12.  Interpret  each,  of  the  parts  of  Ex.  11  as  a  verbal  problem. 

13.  Through  how  many  degrees  must  the  line  OI^  turn 
(Fig.  155)  to  reach  the  position  OR2? 

14.  Archimedes,  a  great  mathema- 
tician, was  born  about  the  year  —  287 
and  was   slain   by  a    Roman    soldier 
in  —  212  while  studying  a  geometrical 

figure  that  he  had  drawn  in  the  sand.  FIG.  155 

How  old  was  he  ? 

15.  Livy,  a  famous  Roman  historian,  was  born  in  —  59  and 
lived  to  be  76  yr.  old.    In  what  year  did  he  die  ? 

16.  Herodotus,  the  Greek  historian,  sometimes  called  the 
Father  of  History,  was  born  in  —  484  and  died  in  —  424.    At 
what  age  did  he  die  ? 

220.  Subtraction   illustrated  by  the  number  scale.    In 
subtracting  4  from  6  we  find  what  number  must  be  added 

-8  -7  -6  -5  -4  -3  -2  -1     0  +1  4-2  +3  4-4  -H5  4-6  +7  4-8 
< — i — I — I — I — I — I — I — I — I — I — I — I — I — i — i — i — i — > 

FIG.  156.   THE  NUMBER  SCALE 

to  the  4  (the  subtrahend)  to  get  the  6  (the  minuend). 
On  the  number  scale  (Fig.  156)  how  many  spaces  (begin- 
ning at  4)  must  we  count  until  we  arrive  at  6  ? 


POSITIVE  AND  NEGATIVE  NUMBEKS        171 

ILLUSTRATIVE  EXAMPLES 

1.  Subtract  —  2  from  3. 

Solution.  Beginning  at  —  2  we  need  to  count  5  spaces  to  the 
right  (positive)  to  arrive  at  3.  Hence,  subtracting  —  2  from  3  equals  5. 
Note  that  we  might  have  obtained  the  result  by  adding  +  2  to  3. 

2.  Subtract  +  5  from  —  2. 

Solution.  Beginning  at  5  on  the  number  scale  we  need  to  count  7 
to  the  left  (negative)  to  arrive  at  —  2.  Hence,  subtracting  +  5  from 
—  2  equals  —  7.  Note  that  we  could  have  obtained  the  same  result 
by  adding  —  5  to  —  2. 

This  exercise  may  be  stated  as  a  temperature  problem ;  namely, 
What  is  the  difference  between  5  above  zero  and  2  below  zero  ? 

3.  Subtract  —  8  from  —  2.    Interpret  as  a  verbal  problem. 

Solution.  Beginning  at  —  8  on  the  number  scale  we  need  to  count 
6  to  the  right  (positive).  Hence,  subtracting  —  8  from  —  2  equals  +  6. 
Notice  that  the  same  result  is  obtained  if  +  8  is  added  to  —  2. 

These  examples  show  that  since  subtraction  is  the  reverse 
of  addition,  we  can  subtract  a  number  by  adding  its  opposite. 
Thus,  adding  $100  to  the  unnecessary  expenses  of  a  firm 
is  precisely  the  same  as  subtracting  $100  gain,  or,  on 
the  other  hand,  eliminating  (subtracting)  $1000  of  lost 
motion  in  an  industrial  enterprise  is  adding  $1000  to  the 
net  gain. 

It  is  convenient  for  us  to  make  use  of  this  relation,  for 
by  its  use  there  will  be  no  new  rules  to  learn,  but  merely 
an  automatic  change  of  sign  when  we  come  to  a  subtraction 
problem,  and  a  continuation  of  the  process  of  addition. 

221.  Algebraic  subtraction.  The  preceding  discussion 
shows  that  subtraction  of  algebraic  numbers  may  be  changed 
into  algebraic  addition  by  the  following  law:  To  subtract 
one  number  from  another  change  the  sign  of  the  subtrahend 
and  add  the  result  to  the  minuend. 


172  GENERAL  MATHEMATICS 

Thus,  the  subtraction  example 

+    7a 

-    3a 

+  10  a 

may  be  changed  to  the  addition  example 

+  la 
+  3a 
+  10  a 

EXERCISES 

Subtract  the  lower  number  from  the  upper  number  in  the 
following.    Illustrate  Exs.  1-11  with  verbal  problems. 


1. 

29 

9.  -14 

17.  +x 

25. 

+  0.81  cc2 

-10 

-    3 

±x 

f 

-2.62x2 

2. 

-55 

10.   +14 

18.   —x 

26. 

-.660 

-15 

+    3 

+  # 

-.840 

3. 

-65 

11-   -T9o 

19.   +x 

27. 

+  0.82  r 

+  15 

.     2 

—  X 

+  2.41  r 

4. 

+  18 

12.  -5  a-2 

20.   +5.74  a;8 

28. 

-  3.34  a 

+  24 

+     a-* 

-6.26  a-8 

+  5.37  a 

5. 

A 

13.   -1268 

21»   -3.15  a2 

29. 

+  2.04y 

T2 

7b3 

.34  a2 

-4.23y 

6. 

3 
"ff 

14.   -      9r 

22.   +6(a  +  &) 

30. 

+  8.92  a; 

~  T 

-  14  /• 

—  8(a  +  J) 

+  9.17  -jr 

7. 

3 

15.   +    7«2 

23.  _5(a-6) 

31. 

-  7.42  z 

14 

-14  a2 

3  (a  —  ft) 

-  3.71  « 

8. 

14 

16.   -7m8 

24.   -4.36* 

32. 

-2.417/ 

-    3 

+  3m8 

8.64* 

+  8.62?/ 

POSITIVE  AND  NEGATIVE  NUMBERS        173 

TRANSLATION  INTO  VERBAL  PROBLEMS 

33.  Translate  each  of  the  following  subtraction  exercises 
into  a  verbal  problem,  using  the  suggestion  given  : 

+  8246 

(a)  As  assets  and  liabilities  : 

+  5 

(b)  As  gain  or  loss  : 


I 

(c)  As  debit  or  credit  : 


—  48 
(d)  As  an  angle  problem  : 


-27 

14° 


+  22 
(e)  As  an  age  problem  (time)  : 

(f  )  As  line  segments  on  the  number  scale  : 

—  246 

(g)  As  a  bank  account  : 


—  40 

(h)  As  a  latitude  problem  : 

-(-  Zo 

i  go 
(i)  As  a  longitude  problem: 

~l~  75 

i   £2 
(j)  As  a  problem  involving  forces  : 

222.  Subtraction  of  polynomials.  When  the  subtrahend 
consists  of  more  than  one  term  the  subtraction  may  be 
performed  by  subtracting  each  term  of  the  subtrahend 
from  the  corresponding  term  of  the  minuend. 

For  example,  when  we  wish  to  subtract  5  dollars,  3  quarters,  and 
18  dimes  from  12  dollars,  7  quarters,  and  31  dimes,  we  subtract  5 
dollars  from  12  dollars,  leaving  7  dollars  ;  3  quarters  from  7  quarters, 
leaving  4  quarters  ;  and  18  dimes  from  31  dimes,  leaving  23  dimes. 


174  GENERAL  MATHEMATICS 

The  subtraction  of  algebraic  polynomials  is,  then,  not 

different    from   the   subtraction    of   monomials  and    may 

therefore    be    reduced   to   addition,   as   in   the  following 
two  examples,  which  are  exactly  equivalent: 

SUBTRACTION  ADDITION 

7  «2  -  14  ab  -  11  P  7  a2  -  14  ab  -  11  tf 

+  5  a2  +    3  ab  +    3  ft2  -  5  a2  -    Sab-    3  b* 

2  a*  -  17  ab  -  14  ft2  2  a2  -  17  «//  -  14  ft2 

The  student  should  change  the  signs  of  the  subtrahend 
in  the  written  form  until  there  is  no  doubt  whatever  as 
to  his  ability  to  change  them  mentally.  The  example  will 

appear  as  follows: 

«2  +  2  ab  +  //•» 

-      + 

+  a2  -  2  ab  +  fe2 

4o6 

NOTE.  The  lower  signs  are  the  actual  signs  of  the  subtrahend. 
They  are  neglected  in  the  adding  process. 

Numerous  verbal  problems  have  been  given  with  the 
hope  of  giving  a  reasonable  basis  for  the  law  of  subtraction. 
The  student  should  now  apply  the  law  automatically  in  the 
following  exercises. 

EXERCISES 

Subtract  the  lower  from  the  upper  polynomial  : 

1.  4a?-3ab  +  6b2  3.  x3  +  3x2y  +  Sxy2  +  y3 

4:a?-5ab-4:b2  -  7  xzy  +  3  xy*  +  y* 


2.          x2-5zy+     if  4.       2mn2  +  5msn+    6 

-3x2-4a;//-3y2  -  7  mri*  -  4  m*n  +  18 

5.  From  10  xy  —  5  xz  -j-  6yz  subtract  —  4  xy  —  3  xz  +  3  yz. 

6.  From  16  x»  -  5  mx*  +  4  m*  subtract  7x*  —  4  mx*  +  12  m3. 

7.  From  2a?-2a?b  +  al>2-2b3  subtract  a3  -  3  a?b  +  aft2  -  6s. 


Subtract  as  indicated,  doing  as  much  of  the  work  as  possible 
mentally. 

8.  (4r8  —  6  r8*  +  10  s8  -  6  rs8)  -  (2  r8  +  6  r2*  +  4  s8  -f-  3  rs2). 

9.  (_  8 m*pq - 4 msp -10mV) -(-6m*p-8m?pq - 15  mV)- 

10.  (15 xs  -  12 afy  +  Ty8)  -  (-  11  z8  +  8 a"y -  5  y8). 

11.  (£  a8  -  3J  aft8  -  3  a8*)  -  (-  f  a&2  +  5^  «26  -  3  a8). 

12 .  (5f  rst  -  7  J  r8*  -  8f  s8*  -  6 J  *8/-)  -  (4 J  s2*  +  3|  »•**  +  7 J  A). 

13.  (2.3 aW - 4.6  a468  +  8.7a6*2) -(-1.1  aW -  2.1  a6*8 - 3a364). 

14 .  (3  x2  -  4  x  +  5)  -f  (2  x2  +  5  x  -  3)  -  (-  2  a2  +  3  x  +  6). 

15.  (5.2  ofy  -  41xy  +  2  if)  -  (31  afy  +  3.2  xy  +  5  y2). 

16.  (2.42  a262  +  5  ab  +  6)  -  (3.12  a8^—  2  ai  -  9). 

17.  (3  a63-  3  afe8)  -  (-2  a&8  +  3  a3- 4 a5c8)  -  (-  4 a8- 

18.  (5x2+2a-//  +  3y2)  +  (2*2-5a;y-/)-(9x2 

19.  Compare  the  signs. of  the  terms  of  the  subtrahend  in  the 
foregoing  exercises  before  and  after  the  parenthesis  are  removed. 

20.  State  a  rule  as  to  the  effect  of  the  minus  sign  preceding 
a  polynomial  in  parenthesis. 

21.  What  is  the  rule  when  the  plus  sign  precedes  a  poly- 
nomial in  parenthesis  ? 

223.  Symbols  of  aggregation.  It  has  been  found  very 
convenient  to  use  the  parenthesis  for  grouping  numbers. 
Such  a  symbol  indicates  definitely  where  a  polynomial 
begins  and  ends.  Other  symbols  used  with  exactly  the 
same  meaning  and  purpose  are  [  ]  (brackets)  ;  {  }  (braces)  ; 
and  "  "  (vinculum).  Thus,  to  indicate  that  a  +  b  is 
to  be  subtracted  from  x  +  y  we  may  use  any  one  of 
the  following  ways :  (#  +  #)  —  (#  +  ft),  [x  +  y\  —  [a  +  ft]. 
{x-}- y}  —  {a  +  ft},  or  x+y  —  a  +  b.  The  vinculum  is  like 
the  familiar  line  separating  numerator  and  denominator 

,.       .       .  .2        a  +  ft 

oi  a  fraction,  as  in  -  or  -•• 

o        a—  ft 


170'  (JEMERAL  MATHEMATICS 

Sometimes  the  symbols  are  inclosed  one  pair  within 
another;  thus,  19  -  (16  -  (9  -  2)}. 

In  an  example  like  the  preceding  the  common  agree- 
ment is  to  remove  first  the  innermost  parenthesis.  First, 
2  is  to  be  subtracted  from  9,  then  the  result,  7,  is  to  be 
subtracted  from  16.  This  result,  9,  is  in  turn  to  be  sub- 
tracted from  19 ;  whence  the  final  result  is  10. 

EXERCISES 

1.  (live  the  meaning  of  the  following: 
(a)  15 -{4 +  (6 -8)}. 


(C)  _  5 x  _  [_  7 x  _  {—2x- 
(d)  3 (a;  +  y) -  5{x  -  2x-3y}. 

2.  Keep  definitely  in  mind  the  rules  governing  the  effect  of 
a  minus  or  a  plus  sign  before  a  grouping  symbol.  Perform  the 
following  indicated  operations  and  simplify  the  results : 

(a)  12-{5-(-2x-5)}. 

(b)  17  -  {-  12  x  -  3  x  -  4}. 

(c)  4  a2  -  (a2  -  3  a8  +  3  a2  -  a8}. 

(d)  2e-[6e-36-4e-(2e-46)]. 

(f )  15  a?  -  {-  3  x2  -  (3  x2  +  5)}  -  (20  a2  +  5). 

SUMMARY 

224.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  positive  number,  negative  number, 
algebraic  numbers,  absolute  value  of  a  number  (or  numerical 
value),  degree  -of  a  number,  degree  of  a  monomial,  degree 
of  a  polynomial,  descending  power,  ascending  power. 


POSITIVE  AND  NEGATIVE  NUMBEKS         177 

225.  The  following  symbols  were  used :   -f  (plus  sign) 
and  —  (minus  sign)  for  positive  and  negative  number  re- 
spectively, (  )  (parenthesis),  [  ]  (brackets),  {  }  (braces), 
and  ~~    ~  (vinculum). 

226.  Positive  and  negative  numbers  have  been  illustrated 
by  game  scores,  directed  line  segments,  latitude,  longitude, 
time,  the  number  scale,  forces,  the  thermometer,  angles, 
profit  and  loss,    debit  and    credit,   assets  and    liabilities, 
deposits   and  checks. 

227.  The  sum  of  two  algebraic  numbers  with  like  signs 
is  the  sum  of  their  absolute  values  with  their  common 
sign  prefixed. 

The  sum  of  two  algebraic  numbers  with  unlike  signs 
equals  the  difference  of  their  absolute  values  with  the  sign 
of  the  number  having  the  greater  absolute  value  prefixed. 

The  sum  of  three  or  more  monomials  is  found  most 
easily  by  the  following  method :  (1)  add  all  positive  terms, 
(2)  add  all  negative  terms,  (3)  add  the  two  sums  obtained. 

228.  To  add  polynomials  add  the  similar  terms  (write 
similar  terms  in  the  same  column). 

229.  To  find  the  difference  of  two  numbers  change  the 
sign  of  the  subtrahend  and  add. 

230.  A  parenthesis  is  used  for  grouping.    If  preceded 
by  a  plus  sign,  it  may  be  removed  without  making  any  other 
changes ;  if  preceded  by  a  minus  sign,  it  may  be  removed 
if  the  sign  of  every  term  within  the  parenthesis  is  changed. 


CHAPTER  IX 


+  4 


POSITIVE  AND  NEGATIVE  NUMBERS.    MULTIPLICATION 
AND  DIVISION.     FACTORING 

231.  Multiplication.  The  laws  of  multiplication  of  num- 
bers having  plus  or  minus  signs  are  easily  applied  to  a 
considerable  number  of  interesting  problems.  These  laws 
are  illustrated  in  the  following  examples: 

ILLUSTRATIVE  EXAMPLES 

1.  Find  the  product  of  (+  4)  and  (+  2). 

Solution.  Geometrically  we  interpret  this  as  follows :  Take  a 
segment  +  4  units  long  and  lay  it  off  two  times  to  the  right  of 
zero  on  the  number  scale ;  that 
is,  in  its  ewn  direction  (Fig.  157). 
Thus,  (+2) (+4)  =  +  8. 

2.  Find    the    product    of 
(-4)  and  (+2). 

Solution,  Geometrically  this 
means :  Take  a  segment  —4  units 
long  and  lay  it  off  two  times  to 
the  left  of  zero  on  the  number 
scale ;  that  is,  in  its  own  direction 
(Fig.158).  Thus,(  +  2)(-4)  =  -8. 

3.  Find    the    product    of 
(+  4)  and  (-  2). 

Solution.  Geometrically  we  in- 
terpret this  as  follows:  Take  a 
segment  +  4  units  long  and  lay  it  off  two  times  to  the  left  of  zero ; 
that  is,  opposite  its  own  direction  (Fig.  159).  Thus,  (—  2)  (+  4)  =  —  8. 


i  i  i   i   i   i   i   i   i    i  i   i   i   i   i  i 
+  8       r^  X 


-4 


FIG.  157 


—4 


L  -4 


O 
FIG.  158 

-4     J 


I    I    I    I    i    I 


O 
FIG.  159 


POSITIVE  AND  NEGATIVE  NUMBEKS        179 

4.  Find  the  product  of  (-  4)  and  (-  2). 

Solution.  If  the  first  factor  were  a  positive  2,  then  we  should 
interpret  this  geometrically  by  laying  off  —  4  twice,  obtaining  a 
line  segment  —  8  units  long  (see 
ORV  in  Fig.  160)  just  as  we  did  in 
Ex.  2.  But  since  it  is  a  negative  2, 
we  lay  it  off  not  in  the  direction  of 
ORl  but  in  the  opposite  direction;  p  , 

namely,  OR  (see  Fig.  161).    Thus, 


Note    that    in 
this  last  case,  as 

in  Ex.1,  the  signs 

°  FIG.  161 

of   the   multipli- 
cand and  the  multiplier  are  alike,  and  the  product  is  positive  ; 
while  in  Exs.  2  and  3  the  signs  of  the  multiplicand  and 
multiplier  are  unlike,  and  the  product  is  negative. 

EXERCISES 

1.  Find  geometrically  the  products  of  (+  2)  (+  5);  (-  2)(+5); 
(+2)  (-5);  (-2)  (-5). 

2.  State  the  law  of  signs  for  the  product  of  two  algebraic 
numbers  as  suggested  by  the  preceding  work. 

232.  The  law  of  signs  for  multiplication.     The  law  of 

signs  for  multiplication  is  as  follows: 

The  product  of  two  factors  having  like  signs  is  positive. 
The  product  of  two  factors  having  unlike  signs  is  negative. 

EXERCISES 

Find  the  value  of  the  following  products,  using  the  law  of 
signs.    Illustrate  the  first  ten  geometrically. 

1.  (+3)  (+5).  3.  (-3)  (+5).  5.  (-2)  (+3). 

2.  (-  3)(-  5).  4.  (+  3)(-  5).  6.  (-  2)(-  3). 


180 


GENERAL  MATHEMATICS 


7.  (-2)  (+7). 

8.  (+2)  (+7). 

9.  (9)  (-3). 
10.  (-4) (-a). 


11.  (2)(-«). 

12.  (-!)(-!). 

13.  (-2*) (-3). 

14.  2a/>-3. 


15.  (-3)  (-5  a). 

16.  (-f)( 

17.  (-f)( 

18.  -2o- 


FIG.  162.    THE  LAW  or  MULTI- 
PLICATION   ILLUSTRATED   BY  THE 
BALANCED  BEAM 


233 !.  Law  of  multiplication  illustrated  by  the  balance. 
The  law  of  signs  may  be  illustrated  with  a  balanced  bar 
(Fig.  162).  A  light  bar  is  balanced  at  M.  The  points  rv  r2, 
etc.  represent  pegs  or  small 
nails  driven  at  equal  distances. 
We  shall  speak  of  rv  rv  etc. 
as  "  first  right  peg,"  "  second 
right  peg,"  etc.  and  of  lv  12, 
etc.  as  "  first  left  peg,"  "  sec- 
ond left  peg,"  etc.  with  the  bar 
in  a  position  facing  the  class  as 
in  Fig.  162.  The  weights,  w, 
are  all  equal ;  hence  we  shall 
merely  speak  of  them  as  "  two  weights,"  "  three  weights," 
etc.  instead  of  mentioning  the  number  of  ounces  or  grams 
contained.  In  Experiments  1-3  the  string  over  the  pulley 
is  fastened  on  the  first  left  peg. 

EXPERIMENTS 

1.  Hang  two  weights  on  /r  This  tends  to  turn  the  bar.  How 
many  must  be  attached  to  the  hook  H  to  keep  the  bar  level  ?  Hang 
three  weights  on  /r  What  do  you  notice  about  the  turning  tendency 
as  compared  with  the  first  case  ?  Answer  the  same  question  for  four 
weights  on  13. 

1  The  entire  article  may  be  omitted  at  the  teacher's  discretion.  The 
device  has,  however,  proved  useful  in  the  hands  of  many  teachers.  The 
apparatus  may  be  bought  at  several  of  the  large  book  companies  or, 
better  still,  made  in  the  shop  by  a  member  of  the  class,  using  a  part  of 
a  yardstick  for  the  lever  and  small  nails  for  pegs. 


POSITIVE  AND  NEGATIVE  NUMBERS        181 

2.  Hang  one  weight  on  lr   How  many  must  be  placed  on  the  hook 
to  keep  the  bar  level?  Hang  one  weight  on  /2 ;  remove  it  and  hang 
one  weight  on  /3 ;  on  /4 ;  and  so  on.    What  do  you  notice  about  the 
turning  tendency  in  each  case  ?    What  two  things  does  the  turning 
tendency  seem  to  depend  on  ? 

3.  With  the  string  passing  over  the  pulley  fastened  to  /j  ho.w 
many  weights  must  be  put  on  the  hook  to  balance  two  weights 
on  /8  ?  three  weights  on  12  ?  one  weight  on  lt  ?  two  weights  on  Z4  ? 
three  weights  on  /4? 

4.  Repeat  Experiments  1~3  for  the  pegs  on  the  right  side,  with  the 
pulley  string  fastened  to  rr   What  seems  to  be  the  only  difference? 

Results  of  experiments.    The  experiments  show  that 

1.  The  turning  tendency  (Jorce)  varies  as  the  number  of 
weights  hung  on  a  peg  on  the  bar.    Thus,  the  more  weights 
hung  on  any  peg,  the  stronger  the  force. 

2.  The  turning  tendency  also   varies  as  the  distance  of 
the  peg  from  the  turning  point. 

3.  The   turning  tendency  is  equal  to  the  product  of  the 
iveights  multiplied  by  the  distance  of  the  peg  on  which  the 
weight  hangs  from  the  turning  point. 

4.  When  a  weight  is  hung  on  a  right  peg,  the  bar  turns  in 
the  same  direction  as  the  hands  of  a  clock ;  when  a  weight  is 
hung  on  a  left  peg,  the  bar  rotates  in  a  direction  opposite  to 
the  hands  of  a  clock. 

234.  Signs  of  turning  tendency ;  weight ;  lever  arm.  It 
is  conventionally  agreed  that  when  the  bar  turns  counter- 
clockwise (as  you  face  it),  the  turning  tendency  is  positive ; 
while  if  the  bar  rotates  clockwise,  the  turning  tendency  is 
negative. 

Weights  attached  to  the  pegs  are  downward-pulling 
weights  and  are  designated  by  the  minus  sign.  Weights 
attached  at  H  pull  upward  on  the  bar  and  are  designated 
by  the  plus  sign. 


182  GENERAL  MATHEMATICS 

The  distance  from  the  turning  point  to  the  peg  where 
the  weight,  or  force,  acts  will  be  called  the  lever  arm,  or 
•arm  of  the  force.  Lever  arms  measured  from  the  turning 
point  toward  the  right  will  be  marked  + ;  those  toward 
the  left,  — .  For  example,  if  the  distance  from  M  to  peg  i\ 
is  represented  by  +  1,  then  the  distance  from  M  to  r3  will 
be  +  3 ;  the  distance  from  M  to  /2  will  be  —  2  ;  and  so  on. 

235.  Multiplication  of  positive  and  negative  numbers.  By 
means  of  the  apparatus  (Fig.  162)  the  product  of  positive 
and  negative  numbers  is  now  to  be  found. 

ILLUSTRATIVE  EXAMPLES 

1.  Find  the  product  of  (+  2)  (-  4). 

Solution.  AVe  may  interpret  this  exercise  as  meaning,  Hang  four 
downward-pulling,  or  negative,  weights  on  the  second  peg  to  the 
right  (positive).  The  bar  turns  clockwise.  The  force  is  negative; 
hence  the  product  of  (2)  (—  4)  is  —  8. 

2.  Find  the  product  of  (-  2)  (-  4). 

Solution.  Hang  two  downward-pulling,  or  negative,  weights  on  the 
fourth  peg  to  the  left  (negative).  The  bar  turns  counterclockwise. 
The  force  is  positive ;  hence  the  product  of  (—  2)  (—  4)  is  +  s. 

3.  Show  that  (+  3)  (+  4)  =  +  12. 

HINT.  Fasten  the  string  over  the  pulley  to  the  fourth  peg  to  the 
right  and  hang  three  weights  on  the  hook. 

4.  Show 'that  (-3)  (+2)  =  -6;  that  (+ 2) (- 3)  =  - 6.    How 
does  the  beam  illustrate  the  law  of  order  in  multiplication  ? 

5.  Compare  the  results  of  Exs.  1-4  with  the  law  of  signs  in 
multiplication  (Art.  232). 

It  is  hoped  that  the  law  of  signs  is  made  reasonablv  clear 
by  means  of  these  illustrations.  The  student  should  now 
proceed  to  ajjply  the  law  automatically. 


POSITIVE  AND  NEGATIVE  NUMBERS        183 


EXERCISES 


State  the  products  of  the  following,  doing  mentally  as  much 
of  the  work  as  possible  : 


1.  (+4)  (-6). 

2.  (-4)  (+6). 

3.  (+4)  (+6). 

4.  (-4)  (-6). 

5.  (+2)  (+5). 

6.  (+3)  (-4). 

7.  (-5)  (-2). 

8.  (-3)  (-7). 

9.  (-5)  (+6). 


11.  (-3.1)  (-5).        21. 

12-  (-f)(f).  22.  (-6X-S). 

23.  (-8)(- 

24.  (-c)(- 

25.  (- 

26.  (- 

17.  (+6j-)(+6j).      27.  (51) (-^2) 

18".  (+6j)(-6-i).      28.  (-9)(+x2 


16.     _6- 


19.  (_6i)(+6-i-). 


10.  (-12)  (-13).   20. 


29.  (-1)3. 

30.  -23. 


236.  Multiplication  by  zero.   The  product  of  3  x  0  means 
0  +  0  +  0  =  0. 

EXERCISES 

1.  Show  geometrically  that  ax  0=0. 

2.  Show  by  the  beam  (Fig.  162)  that  a  x  0=0;  that  0  x  a=0. 

3.  State  a  verbal  problem  in  which  one  of  the  factors  is 
zero.    In  general  both  a  x  0  and  0  x  a  equal  zero.    Hence  the 
value  of  the  product  is  zero  when  one  of  the  factors  is  zero. 

4.  What  is  the   area  of  the  rec- 
tangle in  Fig.  163  ?  How  would  the  area 
change  if  you  were  to  make  the  base 
smaller  and  smaller  ?  What  connection 
has  this  with  the  principle  a  x  0  =  0  ? 

5.  How  would  the  area  of  the  rec- 

tangle in  Fig.  163  change  if  I>  were  not  changed  but  a  were 
made  smaller  and  smaller  ?    What  does  this  illustrate  ? 


b 
FIG.  163 


184  GENERAL  MATHEMATICS 

237.  Product  of  several  factors.  The  product  of  several 
factors  is  obtained  by  multiplying  the  first  factor  by  the 
second,  the  result  by  the  third,  and  so  on.  By  the  law  of 
order  in  multiplication  the  factors  may  first  be  rearranged 
if  this  makes  the  exercise  easier.  This  is  often  the  case  in 
a  problem  which  involves  fractions. 

EXERCISES 

1.  Find  the  value  of  the  following  products  : 
(a)  (+2)  (-3)  (-5)  (-4). 


00  (-  -BX-f)  (-!!)(¥)• 

2.  Find  the  value  of  (-1)2;  (-  1)3;  (-1)4;  (-2)2(-2)3 
(-2)*;  (-2)5;  (-  3)2(-  3)8(-  3)*;  (_4)2(-4)8. 

3  .  Find  the  value  of  3  a-4  —  5  xs  +  x*  —  12  x  —  5  when  x  —  —  2. 

4.  Find   the  value  of  x8  —  3x*y  +  Secy2  +  y*  when   x  =  3 
and  y  =  —  2. 

5.  Find  the  value  of  z8  +  3  x2  +  3  x  +  1  when  x  =  10. 

6.  Compare  (-  2)3  and  -  2s  ;  -  3s  and  (-  3)8  ;  (-  2)4  and 
-  24  ;  (-  a)8  and  -  «8  ;  (-  a)4  and  -  a*. 

7.  What  is  the  sign  of  the  product  of  five  factors  of  which 
three  are  negative  and  two  are  positive?  of  six  factors  of 
which  three  are  negative  and  three  are  positive  ? 

8.  What  powers  of  —  1  are  positive  ?   of  —  2  ?   of  —  x  ? 
State  the  rule. 

238.  Multiplication  of  monomials.    Find  the  product  of 


The  sign  of  the  product  is  determined  as  in  Art.  232  and  is  found 
to  be  +  . 

By  the  law  of  order  in  multiplication  the  factors  may  be  arranged 

as  follows  :  0  ,     QN 

2(-  3)  (-  o)  zxxxyyy, 

which  is  equal  to  3 


POSITIVE  AND  NEGATIVE  NUMBERS         185 

Hence,  to  find  the  product  of  two  or  more  monomials 

1.  Determine  the  sign  of  the  product. 

2.  Find  the  product  of  the  absolute  values  of  the  arith- 
metical factors. 

3.  Find  the  product  of  the  literal  factors. 

4.  Indicate  the  product  of  the  two  products  just  found. 

EXERCISES 

Simplify  the  following  indicated  products,  doing  mentally  as 
much  of  the  work  as  possible  : 


2.  (-2*)  (-3*)  (4*). 

3.  (5  xV2)  (-  3  *V«)  (2  ay*3)  (-  7  ay*2). 

4.  (-  2  a2bcd)  (5  aPcd)  (-  7  abc2d)  (-  2  alcd*). 

5.  (—  3  mnx)  (—  5|-  m*nx)  (—  2  oj). 


7.  (-  3J  06)  (-  5  J  a2^2)  (+  8  oft). 

8-  (5i?V)(l^X 

9.  (I  Vty  (6  jac^^)  ( 

10.  (3  ^)  (6  m2m^}  ( 

11.  (1.3  xV*)(-  2  aVu*u 

12.  (1.1  a-y2)  (1.1  mxy2)  (10  «i2o;2). 

13.  (_2yX-3)2(-4)(2)2. 

14.  (-  3)2(-  2)2(-  I)3. 

15.  (5)(-4)(3)2(0)(2)2. 

16.  (-  a)\-  a}\-  a2)  (-  a3). 

17.  (aj  +  y)(a5  +  y)»(a5  +  y)6. 

18.  3  (aj  +  y)8(x  -  7/)2(x  +  y)'(x  -  y}*. 

19.  Find  the  value  of  1+  (-  5)  (-  3)  -  (-  2)  (4)  -  (-  3)  (3; 


186  GENERAL  MATHEMATICS 

20.  What  is  a  short  method  of  determining  the.  sign  of  the 
product  containing  a  large  number  of  factors  ? 

NOTE.  It  is  agreed  that  when  an  arithmetical  expression  contains 
plus  or  minus  signs  in  connection  with  multiplication  or  division 
signs,  the  multiplication  and  division  shall  be  performed  first.  This 
amounts  to  the  same  thing  as  finding  the  value  of  each  term  and  then 
(tili/in</  in-  s  it  lit  racling  as  indicated. 

239.  Multiplication  of  a  polynomial  by  a  monomial.  We 
shall  now  see  how  the  process  of  algebraic  multiplication 
is  extended. 

INTRODUCTORY  EXERCISES 

1.  Keview  the  process  of  finding  the  product  of  a(x  +  y  +  z) 
in  Art.  122. 

2.  Illustrate  by  a  geometric  drawing  the  meaning  of  the 
product  obtained  in  Ex.  1. 

3.  How  many  parts  does  the  whole  figure  contain  ? 

4.  What  is  the  area  of  each  part  ? 

The  preceding  exercises  serve  to  recall  the  law  that  a 
polynomial  may  be  multiplied  by  a  monomial  by  multiplying 
every  term  of  the  polynomial  by  the  monomial  and  adding  the 
resulting  products. 

DRILL  EXERCISES 

Find  the  products  as  indicated  and  check  by  substituting 
arithmetical  values  for  the  literal  numbers  : 


Solution.  a2  -  2  ab  +  3  b*  =    W 

_  3_a=  _6 
3  a3  -  6  a26  -f  9  a&2  =  114 


Check.  Let  a  =  2  and  b  =  3.  Then  the  same  result  is  obtained  by 
substituting  in  the  product  as  by  substituting  in  the  factors  and 
then  multiplying  the  numbers.  Note  that  the  check  is  not  reliable 


AND  NEGATIVE  NUMBERS        1ST 

if  we  let  a  literal  number  in  a  product  containing  a  power  of  that 
literal  number  (as  a:  in  a  product  a;5  —  4  x)  equal  1,  for  if  x  =  1,  then 
x5  —  4  x  also  equals  x3  —  4  x,  x2  —  4  x,  x9  —  4  x,  etc.  Explain. 

2.  5x(2x*  -3x  -7). 

3. 

4. 

5.   —  5.1  «4(i 

6. 

7.  (?»V  -  3  mV  +  4  mV  -  9  wV)  3.5  ?»V. 

8. 

9. 
10. 
11. 


240.  Product  of  two  polynomials.  In  Art.  126  we  found 
the  product  of  two  polynomials  to  be  the  sum  of  all  the  partial 
products  obtained  by  multiplying  every  term  of  one  polynomial 
by  each  term  of  the  other.  After  reviewing  briefly  the  case 
for  positive  terms  we  shall  proceed  to  interpret  the  above 
law  geometrically  even  when  negative  terms  are  involved. 

ILLUSTRATIVE  EXAMPLES 

1.  Find  the  product  of  (c  +  d)(a  +  It). 

Solution.  The  area  of  the  whole  rectangle  in  Fig.  164  is  expressed 
by  (a  +  &)  (c  +  d).  The  dotted  line  suggests  a  M  d 

method  for  expressing  the  area  as  the  sum  of 
two  rectangles  ;  namely,  a  (c  +  c?)  +  b  (c  +  d). 
If  we  use  the  line  MN,  the  area  may  be 
expressed  as  the  sum  of  four  rectangles  ; 
namely,  ac  +  ad  +  be  +  bd.  Each  expression  c  wo 

equals  the  area  of  one  of  the  rectangles;  hence  FIG.  164 

(a  +  6)  (c  +  d)  =  a(c  +  d)  +  b(c  +  d)  =  ac  +  ad  +  be  +  bd. 


1.88 


GENERAL  MATHEMATICS 


3x  y 

FIG.  165 


G  a  B 


FIG.  166 


2.  Illustrate,    by    means    of    Fig.  165,   the    law    for   the 
multiplication  of  two  polynomials.  %x  y 

3  .  Find  the  product  of  (a  —t>)(c  +  d). 

Solution.  In  this  case  one  of  the  factors 
involves  a  negative  term. 

The  product  (a  —  b)(c  +  d)  is  repre- 
sented by  a  rectangle  having  the  dimen- 
sions (a  -  b)  and  (c  +  rf)  (Fig.  166).  The 
rectangle  ABEF=ac  +  ad.  Subtracting 
from  this  the  rectangles  be  and  bd,  we 
obtain  the  rectangle  A  BCD. 

Therefore  (a  —  b)  (c  +  d)  =  ac  +  ad 
—  be  —  bd,  each  side  of  the  equation  repre- 
senting the  area  of  rectangle  A  BCD. 

*  4  .  Findthe  product  of  (a  —  b)  (a  —  fy  . 

Solution.   Let  ABCD  (Fig.  167),  repre- 
sent a  square  whose  side  is  (a  —  6)  feet. 
Show   that    the    area    of   ABCD    equals 
EFGC  +  GHIB  -  FKDE  -  KHIA. 
Then 

(a  -  ?/)  (a  -  b)  =  (a  -  b)2  Why  ? 

=  a2  +  62  —  ab  —  ab 
=  a2  -  2  ab  +  IP. 

*5.  Sketch  a  rectangle  whose  area  is  (m  +  ri)  (r  —  s)  ;  whose 
area  is  24  b2  —  6  be. 

DRILL  EXERCISES 

Apply  the  law  of  multiplication  to  two  polynomials  in  the 
following  exercises.    Check  only  the  first  five. 


Solution. 


D 

(a-b)' 

A 

B 

b1 

f 

:             G 

1 

FIG.  167 

xz  +  2  xy  +  y- 
x  +  y  _ 
x3  +  2  x'2y  +     xy2 


+  2  xy*  +  y* 


x*  +  3  x*y  +  3  xy*  +  y* 
Check  by  letting  x  =  2  and  y  =  3. 


POSITIVE  AND  NEGATIVE  NUMBERS         189 

2.  (rs  +  tm)  (rs  -  tm).  7.  (2  a  +  3  ft)  (2  a  -  3  4). 

3.  (a8  +  ax3  +  as)  (a  +  a).  8.  (i  aft  -  £  ftc)  (f  aft  +  f  ftc). 

4.  (a2  +  4  a;  +  3)  (a  -2).  9.  (a  +  6  -  c)2. 

5.  (cc2-3cc  +  5)(2x  +  3).  10.  (a  -  b  +  c  -  df. 

6.  (k2  +  3  Ar  +  1)  (A  -  2).  11.  .  (-  2  a  +  3  1>  -  4  c)2. 

12.  Comment    on   the    interesting   form   of  the    results   in 
Exs.  9-11. 

13.  (0.3  a  +  0.4  b  -  0.5  c)  (10  a  -  30  b  +  40  c). 

14.  (2  if  -  12  zy  +  5  x2)  (2  if  -  5  .r2). 

15.  o2  +  *z/  +  z/2)  (a  -  y)  0*  +  y;. 

16.  (9x2 

17.  (x  + 

18.  Comment  on  the  form  of  the  results  in  Ex.  17. 

19.  (r2  +  rs  -  s2)  (r2  +  rs  +  .<?). 

20.  (Sr2  +  5  r  +  6)  (3  r2  +  3  .*  -  6). 

21.  (3x  +  2.y)3-(3x-2y)8. 

22.  (3  x2  +  T/2)2  -  (3  xz  -  iff  -±x*(x- 

23.  (2a-3^)2-(2a  +  36)2  +  (2« 

24.  (0.3  a  -  0.4  4)3  -  (0.5  a  +  0.6  i)2 

-  (0.3  a  -  0.4  ft)  (0.3  a  +  0.4  ft). 


26.  (5  +  3)2-  (5  -  3)2-  (5  +  3)  (5  -  3). 

27.  Why  may  3527  be  written  3  •  10s  +  5  •  10a  +  2  .  1  0  +  7  ? 

28.  Multiply  352  by  243. 

HINT.    Write  in  the  form     3  •  103  +  5  •  10  +  L> 
2  •  102  +  4  •  10  +  3 

29.  Write  56,872  as  a  polynomial  arranged  according  to  the 
descending  powers  of  10. 

30.  Find  the  product  of  5  and  3427  by  the  method  suggested 
in  Ex.  28. 


190  GENERAL  MATHEMATICS     , 

241.  Product  of  two  binomials.  We  shall  now  see  how 
the  algebraic  product  of  two  binomials  may  be  obtained 
automatically.  The  following  exercises  will  help  the  stu- 
dent to  discover  and  understand  the  method. 

EXERCISES 
Find  by  actual  multiplication  the  following  products  : 

1.  (2x  +  3)(4z  +  5).  4.  (4x  +  6)  (4  a;  +  5). 
Solution.   2*  +  3  5.  (3*  -2)  (3*  -2). 

4x+5  6.  (x  +  2)  (*  +  9). 

8*2  +  12*  7.   (2a;+l)(aJ4-6). 

_  +l°*  +  lr'  8.  (b  +  3)  (6  +  5). 

8  x2  +  22  x  +  15 

9.  (a  -7)  (a  -3). 

2.  (3«  +  5)(2a-8).  10'  (3  *  +  8)  (a  +  2). 

11.  (3x  +  4)(2a;-3). 
Solution.    3  a  +  5 

2a-8  12'   (*-3)(*-10). 

13.   (a-8  -  9)  (*"  +  9). 


-  24  a  -  40  14.  (or2  -  5)  (aj2  +  10). 

6a2_14a_40  15>  (3  a;  -5)  (4  a;  -2). 

16.  (2y  —  3)(5y  —  8). 
3.  (o  ?/  +  4)  (o  ?/  —  4). 

17.  (»i» 
Solution,    o  M  +  4                                  1  o  / 


-My  -16  20.  (3a;  +  4y)(3a;-'4y). 

-16  21.  (4a  +  26)(7a-5i). 

22.  53  x  57. 
Solution.    53  x  57  =  (50  +  3)  (50  +  7) 

=  503  +  (7  +  3)50  +  21. 

23.  61  x  69.  24.  52  x  56.      '  25.  37  x  33. 

26.  Can  you  see  any  way  of  formulating  a  rule  for  finding 
the  products  of  two  binomials  ? 


POSITIVE  AND  NEGATIVE  NUMBERS         191 

If  we  agree  to  use  the  binomials  ax  4-  b  and  ex  4-  d  to 
represent  any  two  binomials  where  a,  b,  e,  and  d  are 
known  numbers  like  those  in  the  products  above,  then  we 
may  discover  a  short  cut  in  multiplying  ax  4-  b  by  ex  4-  d. 

ILLUSTRATIVE  EXAMPLE 

Find  the  product  of  (ax  4-  V)  and  (<-x  -\-  <Z). 
Solution.  ax  +  b 

Y 

CX  +  d 


bcx 
.  +  adx  +  bd 


acx2  +  (be  +  ad  )  x  +  bd 

The  arrows  show  the  cross-multiplications  or  cross-products  whose 
sum  is  equal  to  the  middle  term.  It  is  seen  that  the  first  term  of  the 
product  is  the  product  of  the  first  terms  of  the  binomials,  that  the  last  term 
is  the  product  of  the  last  terms  of  the  binomials,  and  that  the  middle  term 
is  the  sum  of  the  tivo  cross-products. 

EXERCISES 

Using  the  rule  stated  above,  give  the  products  of  the  fol- 
lowing binomials  : 

1.  (2  a  +  3)  (3  a  4-  5).  6.  (3  a  -  2fi)(3a  -  2  ft). 
Solution.    The  product  of  the  7.  fx  _  7)  (4  x  +  9). 

first  terms  of  the  two  binomials 

is  6  a2,  the  product  of  the  last 

terms  is  15,  and  the  sum  of  the  9.   (x  +  *>)(x  +  8). 

cross-products  is  19  a.    Therefore  ,„          ^  /7          9v. 

the  product  is  6  a»  +  19  a  +  15. 

2.  (4,,  +  3)  (2.  +  1).  11.  (4,  +  3)(3,-4). 

3.  (2s  -7)  (3*  +  2).  12-  (*4-9&)(aj  +  7ft). 

4.  (3  x  4-  4)  (3x4-  4).  13.  (2  aj  +  4  y)  (3  x  -  5  y). 

5.  (7  x  -  2)  (7  x  +  2).  14.  (5  a  +  4)  (4  a  -  2). 


192  GENERAL  MATHEMATICS 

15.  (7a  +  26)(7o-2i).  18.  (3  a  -  7  ft) (3  a  -  7  6). 

16.  (5a  +  4i)(5«  +  46).  19.  (6  xy  +  2)  (3  xy  -  5). 

17.  (3 a: +  2) (12 a: -20).  20.  (7  ab  +  5c)  (60*  -  8c). 

21.  Do  you  notice  anything  especially  significant  about  the 
product  of  two  binomials  that  are  exactly  alike?    Explain 
by  using  the  product  of  x  +  y  and  a-  +  y,  a  —  b  and  a  —  b 
(compare  with  Ex.  1,  Art.  127). 

22.  Do  you  notice  anything  especially  significant  about  the 
product  of  two  binomials  that  are  the  same  except  for  the 
signs  between  the  two  terms  ?   Explain  by  using  the  product 
of  ra  +  n  and  m  —  n. 

23.  Try  to  formulate  a  rule  for  obtaining  automatically  the 
products  referred  to  in  Exs.  21  and  22. 

242.  Special  products.  We  have  seen  in  Art.  241  how 
the  multiplication  of  two  binomials  may  be  performed 
automatically.  Such  products  are  called  special  products. 
The  student  should  observe  thaf,  Exs.  21  and  22,  Art.  241, 
furnish  examples  of  such  products.  For  example,  the 
product  of  x'+y  and  x  -+-  y  is  equal  to  x2  +  2  xy  +  y*,  and 
is  called  the  square  of  the  sum  of  x  and  y\  while  the 
product  of  a  —  b  and  a  —  b  is  equal  to  a2  —  2  db  +  62,  and 
is  called  the  square  of  the  difference  of  a  and  b.  Further, 
the  product  of  m  +  n  and  m  —  n  is  equal  to  mz  —  n*,  and 
is  called  the  product  of  the  sum  and  difference  of  m  and  n. 

EXERCISES 

1.  Find  automatically  the  following  special  products  and 
classify  each: 

(a)  (x  +  3)(x  +  3).  (c)   (2x  +  4)  (2  a;  +  4). 

(b)  (y_2)(y-2).  (d)  (3* -6)  (3* -6). 

(e)  (2x  +  4)2.  (g)   (2  x  +  4  y)2.  (i)   (2  a  +  4  ft)». 

(f )  (4  x  -  2)2.  (h)  (5  x  -  2  y}\  (j)  (3  a  -  2  £)2. 


193 

2.  State  a  rule  for  finding  automatically  the  square  of  the 
sum  of  two  numbers. 

3.  State  a  rule  for  finding  automatically  the  square  of  the 
difference  of  two  numbers. 

The  preceding  exercises  should  establish  the  following 
short  cuts  for  finding  the.  square  of  the  sum  and  the  square 
of  the  difference  of  two  numbers: 

1.  The  square  of  the  sum  of  two  numbers  is  equal  to  the 
square  of  the  first  number  increased  by  two  times  the  product 
of  the  first  number  and  the  second  number  plus  the  square  of 
the  second  number.    Thus,  (a  -f-  ft)2  =  a2  +  2  ab  +  b2. 

2.  The  square  of  the  difference  of  two  numbers  is  equal 
to  the  square  of  the  first  number  decreased  by  two  times  the 
product  of  the  first  number  and  the  second  number  plus  the 
square  of  the  second  number.    Thus,  (a  —  6)2  =  a2  —  2  ab  -+-  62. 

EXERCISES 
1.  Find  automatically  the  following  products  : 

(a)  (x  +  5)  (x  -  5).  (d)  (c  +  7)  (c  -  7). 

(b)  (a  +  4)  (a  -  4).  (e)  (y  -  11)  (y  +.11). 

(f)  (,.  +  8)(r-8). 


2.  Study  the  form  of  the  results  in  Ex.  1.   Give  the  products 
in  the  following  list  at  sight  : 

(a)  (y  -  10)  (y  +  10).  (d)  (5  -x)(5+  x). 

(b)  (2  x  +  5)  (2  x  -  5).  (e)  (y  -  J)  (//  +  £). 

(c)  (3d-4)(3<*  +  4).  (f)  (5x-y)(5x  +  y). 

3.  Write  the  sum  of  x  and  y  ;  the  difference  ;  find  the  prod- 
uct of  the  sum  and  the  difference.    Check  by  multiplication. 

4.  State  the  rule  for  finding  the  product  of  the  sum  and  the 
difference  of  two  numbers. 


194  GENERAL  MATHEMATICS 

The  preceding  exercises  should  establish  the  following 
short  cut  for  finding  the  product  of  the  sum  and  the 
difference  of  two  numbers: 

1.  Square  each  of  the  numbers. 

2.  Subtract  the  second  square  from  the  first. 


DRILL  EXERCISES 

Find  the  following  products  mentally  : 

1.  (x  +  2)  (aj  +  2).  11.  (x  -  1)  (a;  +  1)  (*2  +  1). 

2-  (U  +  3)  (y  -  3).  12.  (w  -  c)  (w  +  c)  (w*  -  c2). 

3.  («-4)(s-4).  13.-  (10  a;  +  9)  (10  a;  +  9). 

4.  (2w-5)(2tt;  +  6).  14.  (y*if  -  0.5)  (a;2/  +  0.5). 

5.  («  -  2  i)  (s  +  2  J).  15.  (11  +/<7/i2)  (11  +/^2)- 

6.  (3s  +  2o)(3«  +  2a).  16.  (a5  +  ^5)  (a5  -  £5). 

7.  (3r-4«)(3r  +  4«).  17.   (20  +  2)  (20  -  2). 

8.  (Ja  +  j6)(Ja.-ii).  18.  (30  +  1)  (30  -  1). 

9-  (i«y-*)(t*y  +  *)-       19-  (18)(22> 

10.  (x  -  1)  (a  +  1).  20.  (31)  (29). 

243.  Division.  The  law  for  algebraic  division  is  easily 
learned  because  of  the  relation  between  division  and  mul- 
tiplication. ^  We  recall  from  arithmetic  that  division  is  the 
process  of  finding  one  of  two  numbers  when  their  product 
and  the  other  number  are  given  and  also  we  remember 
that  quotient  x  divisor  =  dividend. 

These  facts  suggest  the  law  of  division.  Thus  we  know 
that  +12-=-  +  2  =  -f6  because  (+  2)(+  6)=  +12. 


POSITIVE  AND  NEGATIVE  NUMBERS        195 

EXERCISES 

1.  Since  (2)  (-  6)  =  -  12,  what  is  - 12  -s-  2  ? 

2.  Since  (-  2)  (+  6)  =  -  12,  what  is  —  12  -s-  -  2  ? 

3.  Since  (-  2)  (-  6)  =  +  12,  what  is  +  12  H-  -  2  ? 

4.  Since  (+  «)(+  &)  =  -f  «&,  what  is  (+  aV)  -+•  a? 

5.  If  the  signs  of  dividend  and  divisor  are  alike,  what  is 
the  sign  of  the  quotient  ? 

6.  If  the  signs  of  dividend  and  divisor  are  unlike,  what  is 
the  sign  of  the  quotient  ? 

244.  Law  of  signs  in  division.   The  work  of  the  preced- 
ing article  may  be  summed  up  in  the  following  law:   If 
the  dividend  and  divisor  have  like  signs,  the  quotient  is  posi- 
tive;   if  the    dividend   and   divisor   have   unlike    signs,    the 
quotient  is  negative. 

245.  Dividing  a  monomial  by  a  monomial.  We  shall  now 
have  an  opportunity  to  apply  the  law  learned  in  the  pre- 
ceding article. 

EXERCISES 

Find  the  quotient  in  the  following,  doing  mentally  as  much 
of  the  work  as  possible  : 

1.  (+l5)-(-3)=?  10.  (-10ar)'-5-(-2oj)=? 

2.  (-15)  -L(_3)=?  11-  (-«/>)•*•  (-«)=? 

3.  (-  15)  -t-  (+  3)  =  ?  12-  (-  «•*)  •*•  (+  V)  =  'f 

4.  (+15)-i-(+3)=?  13-  (t *)-«-(- &)  =?- 

5.  (-  18)  -5-  (-  3)  =  ?  14.  (-  0.5  x)  +  (-$x)=  ? 

6.  (-12) -=-(-12)=?  15.  (- 1.21  x2)  ^  (- 1.1  *)  =  ? 

7.  (+5)-f-(+5)=?  16.  (_£)-*.(-£)  =  .?. 

8.  (+*)^(+*)=?  17.  (f)-Kf)=? 

9.  -2a-f.a=?  18'   (-|)^(-f)=? 


196  GENERAL  MATHEMATICS 

19.  $).,.(_•{).?  31.  (-*V(*)=? 

20-  (-1)  ^  (-  1)  =  *  32.  (-  *2)  +  (^>  -  • 

21.  (f)-Kf)='.'  33.  (-9^)-(3«)=? 

22.  (?)  +  (-  t)=-  34'  (-3«*)-l-(-ft)=? 

23.  (_2)  +  (+-J)=?  35.  (6fe)-K-2je)= 

36.  (+«««)  +  (-«)=•.• 


25.   (+12./-)-(-x)  =  ?  38.  <_ 

28.  (4-  «)  -*-  (-  £«)  =  V  39.  <  t-  ww/)  -*-(—  <)  =  ? 

27.  (-*)4-(-J*)=?  40.  (-//Ar8)  -5-(-oA)='.' 

28.  (-  .r*)  -i-  (-  x)  =  ?  41.  (  7  »  «r)  -I-  (-  22)  =  V 

29.  (^  -  (-  x)  =  ?  42.  (oi  r«)  -5-  (-  3j  r)  =  ? 

30.  (  _  x4)  -s-  (-  x)  =  ?  43.  24  a;//  -t-  X  .?  =  ? 

XOTK.  The  algebraic  solution  of  the  more  difficult  problems  of 
this  type  are  best  interpreted  as  fractions,  since  a  fraction  is  an 

•_>4  ,-„ 
indicated  quotient.    Thus,  24  -/•//  -=-3  x  may  be  written  —  —  •    The 

'^  x     24  j-ii 
problem  now  is  one  of  reducing  to  lower  terms.     Thus,  in  - 

both  numerator  and  denominator  may  be  divided  by  3  x.    The  result 

•    8w 

is  —  -  >  or  8  y  units. 

In  algebra,  as  in  arithmetic,  the  quotient  is  not  altered  if  dividend 
aad  divisor  are  both  divided  by  the  same  factor.  Dividing  dividend 
and  divisor  by  the  highest  common  factor  reduces  the  quotient 
(or  fraction)  to  the  simplest  form  (or  to  lowest  terms). 

Solution.    The  sign  of  the  quotient  is  negative.    Why? 
The  numerical  factors  can  be  divided  l>y  S  :  r5  and  x1  are  divisible 
by  x2;  y3  and  »/s  are  divisible  by  y3;  m-  and  w3  are  divisible  by  »?2. 

Hence  ***?#   =  ^  =  -  «f!.- 

—  8  x-im?      —  m  m 


45. 


POSITIVE  AND  NEGATIVE  NUMBERS         197 
-S-jfb  343 


—  xz 


46.^ 
—  9« 


47. 
48. 


49. 
50. 


49  a; 
12 


246.  Dividing  a  polynomial  by  a  monomial.  The  division 
process  will  now  be  extended. 

EXERCISES 

1.  Divide  6  x2  +  4  xij  +  8  xz  by  2  a-. 

As  in  dividing  monomials,  this  quotient  may  be  stated  as  a  rec- 
tangle problem.  Find  the  length  of  the  base  of  a  rectangle  whose 
area  is  6  x"  +  4  xy  +  8  xz  and  whose  altitude  is  2  x.  Indicating  this 
quotient  in  the  form  of  a  fraction, 

we  have 

6  xz  +  4  xy  +  8  xz 


3X 


42 


FIG.  168 


Dividing  numerator  and  denominator 
by  2  x,  the  result  is  3x  +  2 y  +  4~-  Show  that  the  problem  may  now  be  in- 
terpreted by  a  rectangle  formed  by  three  adjacent  rectangles  (Fig.  168). 

2.  Show  that  the  total  area  of  three  adjacent  flower  beds 
(Fig.  169)  may  be  expressed  in  either 

of  the  following  forms  : 

or  5  (3  +  5  +  4). 

Which  form  is  the  better  ?    Why  ? 

3.  Find  the  following  quotients,  obtaining  as  many  as  you 
can  mentally : 

9  a2 -6  a5  .   27«i  +  6«2 


5x3 

5x5 

5x4 

354 

Fir..  169 

(a) 


(g) 


00 


3a 


-4*V 


(e): 
(f) 


6  r/ 


198  GENERAL  MATHEMATICS 

247.  Factoring  ;  prime  numbers.  To  factor  a  number  is 
to  find  two  or  more  numbers  which  when  multiplied 
together  will  produce  the  number.  Thus,  one  may  see  by 
inspection  that  2,  2,  and  3  are  the  factors  of  12.  'In  like 
manner,  x  +  y  and  a  are  the  factors  of  ax  4-  ay. 

A  number  which  has  no  other  factors  except  itself  and 
unity  is  said  to  be  a  prime  number  ;  as,  5,  x,  and  a  +  b. 

A  monomial  is  expressed  in  terms  of  its  prime  factors, 

thus  :  1  5  ax*i?  =  %.5-a-x.x.y.y.y. 

The  following  is  an  example  of  the  method  of  express- 
ing a  factored  polynomial: 


In  algebra,  as  in  arithmetic,  certain  forms  of  number 
expression  occur  very  frequently  either  as  multiplications 
or  as  divisions  —  so  much  so,  that  it  is  of  considerable 
advantage  to  memorize  the  characteristics  of  these  num- 
bers that  we  may  factor  them  by  inspection  and  thus  be 
able  to  perform  the  multiplications  and  divisions  auto- 
matically. In  this  text  we  shall  study  two  general  types 
of  factoring. 

248.  Factoring  Type  I.  Taking  out  a  common  monomial 
factor.  Type  form  ax  +  bx  +  ex  =  x  (a  +  b  +  e). 

A  number  of  this  type  we  shall  call  a  number  containing 
a  common  monomial  factor.  The  products  obtained  in  the 
exercises  of  Art.  239  are  numbers  of  this  type.  Although 
this  type  of  factoring  is  not  difficult,  nevertheless  it  is  im- 
portant and  should  be  kept  in  mind.  We  shall  learn  that 
many  verbal  problems  lead  to  equations  which  can  readily 
be  solved  by  a  method  which  depends  upon  factoring. 
Factoring  also  enables  us  to  transform  formulas  into  their 
most  convenient  form. 


POSITIVE  AND  NEGATIVE  NUMBERS        199 

The  method  of  factoring  this  type  consists  of  the 
following  steps: 

1.  Inspect   the   terms  and   discover   the  factor   which   is 
common  to  all  the  terms. 

2.  Divide  by  the   common   monomial  factor.     The   result 
obtained  is  the  other  factor. 

3.  In  order  to  find  out  whether  he  has  factored  correctly 
the  student  should  multiply  the  two  factors  together. 

NOTE.  In  all  factoring  problems  the  student  should  first  look  to 
see  if  the  number  contains  a  common  monomial  factor. 

EXERCISES 

Factor  the  following  by  inspection  and  check  your  work 
by  multiplication : 

1 .  bx  —  5  b  —  be. 

Solution.  Each  term  has  the  factor  b.  Divide  the  expression  by  b. 
The  quotient  is  x  —  5  —  c. 

Check.  b  (x  —  5  —  c)  =  bx  —  5  b  —  be. 

Therefore  the  factors  of  bx  —  5  6  —  be  are  b  and  x  —  5  —  c. 

2.  5a-5b.  8.  x*  -  x3. 

3.  4cc  +  4?/.  9.  25 x2-  5 x8. 

4.  5xa-lQxb.  10.  2 x2  +  4 xy  +  2 f. 

5.  5  ax2  -  10  axif.  '    11.  d2b  +  ab'2  +  a8. 

6.  2rx8-8?y.  12.  4a;2-8^  +  47/2. 

7.  3  x2  -  6  x.  13 .  a*a?  -  2  aVy2  +  4  aary . 

14.  3  a2  -15  a  +  18. 

249.  Factoring  Type  II.  The  "  cut  and  try  "  method  of 
factoring.  Type  form  acxz+(bc+ad')x+bd=:(az+b')(cz+d'~). 

The  products  obtained  in  the  exercises  of  Art.  241  can  all 
be  factored  easily  by  inspection.  The  method  of  factoring 


200  GENERAL  MATHEMATICS 

such  products  is  known  as  the  "cut  and  try"  or  "trial  and 
error"  method.  The  method  consists  simply  of  guessing 
the  correct  pair  of  factors  from  all  of  the  possible  ones 
and  then  verifying  the  result  by  multiplying  the  factors 
together.  The  method  is  illustrated  by  the  following 
example : 

Factor  2  y?  +  9  ./•  +  10. 

Solution.    There  are  four  possible  pairs  of  factors,  as  shown  below  : 

2  x  +  10  2  x  +  10  x  +  5  2  x  +  .5 

x+    1  2r+    1  -2x  +  2  x  +  2 

It  is  clear  that  the  last  pair  is  the  correct  one,  since  the 
sum  of  the  cross-products  is  9  x.  Of  course  the  correct 
pair  of  factors  may  be  found  at  any  stage  of  the  "cut 
and  try"  method,  and  while  the  process  may  seem  slow  at 
first,  practice  soon  develops  such  skill  that  the  factors  can 
easily  be  found. 

It  is  very  important  for  the  student  always  to  be  sure 
that  the  factors  he  has  obtained  are  prime  numbers.  Such 
factors  are  called  prime  factors.  Incidentally  it  is  impor- 
tant to  remember  that  there  are  some  numbers  that  are 
not  factorable,  because  they  are  already  prime  numbers. 
For  example,  a^+16  and  2^  + 2  a; +12  are  not  factorable. 
See  if  you  can  explain  why  they  are  not  factorable. 

From  what  has  been  said  the  student  will  see  that  in 
all  factoring  problems  it  is  important  to  hold  in  mind 
three  things;  namely: 

1.  Try  to  discover  a  common  monomial  factor. 

2.  Find  the  prime  factors  by  the  "cut  and  try"  method. 

3.  Check  by  multiplying  the  factors  together. 


POSITIVE  AND  NEGATIVE  NUMBERS        201 

EXERCISES 

Find  the  prime  factors  of  the  following  expressions : 
1.  6x*-x-2. 

Solution.  Since  the  x  and  the  2  are  both  negative,  the  last  terms 
of  the  factors  are  opposite  in  sign.  The  possible  combinations  of 
pairs  of  factors  (regardless  of  signs)  are  shown  bclov.  : 

'    Qj;      2  Qx      1  :!./•      2  3x      1 


The  third  pair  is  seen  to  be  correct,  provided  we  write  them 

8  x  -  2  and  2  x  +  1.    Therefore  6  x2  -  x  -  2  -  (3  x  -  2)  (2  x  +  1). 

2.  2 a2  +  9  a  +  9.  23.  2  a-2  -  a-  -  28. 

3.  x2  +  2  xy  +  >/2.  24.  5  x2  -  33  .r  +  18. 

4.  «2  -16.  25.  16 a-6- 25 v/4. 

5.  cc2  +  2x-3.  26.  18 x2  +  21  x  - 15. 

6.  x2  -  2  x  -  3.  27.  6  a2  -  a  -  2. 

7.  4 a;2  +  16.x  +  16.  28.  16-36a-6. 

8.  7  ar  +  9  x  +  2.  29.  a2  -  4  a  -  45. 

9.  9z2-  30  a;  +  25.  30.  2  x2?/  + 11  ccy  + 12  y. 

10.  7/4-s2.  31.  2-2x2. 

11.  x2-5a;  +  4.  32.  24 #d  +  138crf  -  36d. 

12.  5  if"  —  80.  33.  (i~  —  6  cib  —  55  &  . 

13.  6o;2-19x  +  15.  34.  «2//2  -  5  </6  +  6. 

14.  y2  -  5  y  -  6.  35.  100 *V  -  49. 

15.  x2-!.  36.  20x2-./--99. 
IS.  1  —  ??i4  (3  factors).  37.  15  —  4  j-  —  4  a-2. 

17.  3 a-2 -IT a- +10.  38.   18  -  33 x  +  5 a;2. 

18.  «2  -  9  £2.  39.  289  «W  -  81  d*. 

19.  49 -x2.  40.  8  a2 -2. 

20.  3ar  —  4a;+l.  41.  3  a;2  +  4  x  +  2. 

21.  w*a  4- »».  - 1.2.  42.  19x  +  22rK-2- 31. 

22.  5  .r2  -17  x- 12.  43.  .r2  +  20  r  +  84. 


202  GENERAL  MATHEMATICS 

250.  Factoring    perfect    trinomial    squares.     Type  form 
a?  ±  2a6  +  £2  =  O  ±  £)2.    Numbers  like  4z2  +  16z  +  16  or 
x*  —  2  xy  +  y2,  which  are  obtained  by  multiplying  a  bino- 
mial by  itself,  are  called  perfect  trinomial  squares.    They 
are  special  cases  of  the  second  type  of  factoring  discussed 
in  Art.  249.   We  have  already  seen  perfect  trinomial  squares 
where  all  the  terms  are  positive  in  the  problems  of  Ex.  2, 
Art.  127.    See  if  you  can  formulate  a  short  method  of 
factoring  perfect  trinomial  squares. 

EXERCISES 

Factor  the  following  perfect  trinomial  squares  by  a  short 
method : 

1.  a2  +  2ab  +  b2.  5.  9  x2  +  42  xy  +  49  if. 

2.  m2-2mn  +  n2.  6.  64  a2  -32ab  +  ±tf. 

3.  9z2  +  12av/  +  4?/2.  7.  4  x2tf  -  12  xy»  +  9  z*. 

4.  16  a2  -  40  ab  +  25  b2.  8.  9  aty4  +  30  afyV  +  25  «4. 

251.  Factoring  the  difference  of  two  squares.    Type  form 
ai  —  &=(a  +  fr)(a  —  6).  Numbers  of  the  form  a2  —  b2  are 
called  the  difference  of  two  squares.    The  products  obtained 
in  the  exercises  on  page  194  are  numbers  of  this  type. 
This  is  a  special  case  of  the  type  discussed  in  Art.  249. 

ORAL  EXERCISES 

1.  What  is  the  product  of  (x  +  3)  (a;  —  3)?    What  then  are 
the  factors  of  x2  —  9  ? 

2.  State  the  factors  of  the  following : 

(a)  x2  -  4.  (c)  r2  -  4  s2. 

(b)  c2-25.  (d)  25— a* 

3.  Show  by  means  of  Fig.  170  on  the  following  page  that 

a2  -  b2  =  (a  +  ft)  (a  -  b). 


POSITIVE  AND  NEGATIVE  NUMBERS 


208 


The  equation  a2  —  b2  =  (a  —  b)  (a  +  &)  asserts  that  a 
binomial  which  is  the  difference  of  two  squares  may  be 
readily  factored  as  follows  :  M  v 

One  factor  is  the  sum  of 
the  square  roots  of  the  terms 
of  the  binomial,  and  the  other 
the  difference  of  the  square 
roots  of  the  terms  of  the 
binomial. 


FIG.  170 


Thus,  to  factor  49  —  #262 
first  find  the  square  root  of 
each  term ;  that  is,  7  and  ab.  Then,  according  to  the  rule, 
•one  factor  is  7  +  <d>  and  the  other  7  —  ab.  Obviously,  the 
factors  may  be  given  hi  reverse  order.  Why  ? 

EXERCISES 

Factor  the  following  binomials.     Check  by  multiplication 
when  you  are  not  absolutely  certain  the  result  is  correct. 


7.  16  a* -25  6*. 

8.  81rt2-16s2. 

9.  25  a;6 -36s4. 

10.  49  -  36  a-6. 

11.  1- 

12.  \  -a-2. 


1.  a-2 -16. 
2     ,.-1 (,i 

3.  //--I. 

4.  I-./-4. 

5  _      ,2  Q  j  ,2 

6.  9  -  /. 

19.  225  a6  -  wV7tM. 

20.  x4-y4. 

21.  2o?iV-  81m4. 

22.  a-8  —  y8. 

23.  625  «2i4  -  256  a\ 

24.  64  a;6 -9. 

25.  Ca  +  i)a-9. 


13.  100«4a;2-36. 

14.  289m2-  81  ?r. 
15. 

16. 

17.  196 -100  6V. 

18.  361  r2^2  -  196. 

26.  9  (a  +  xf  -16. 

27.  (x3  -  ?/)2  -  x6. 


28. 

29.  0.25«a-0.64Ja. 

30.  0.25  r/2-^- 

lo 


204  GENERAL  MATHEMATICS 

Knowledge  of  the  special  products  considered  above 
enables  us  to  multiply  certain  arithmetic  numbers  with 
great  rapidity.  Thus  the  product  of  32  by  28  may  be 
written  (30  +  2)  (30  -  2)  =  (30)2  _  (2)«  =  896. 

EXERCISES 

1.  Give  mentally  the  following  products : 

(a)  18-22.  (e)  32-27.  (i)  67-73.  (in)  75-85. 

(b)17-23.  (f)37-43.  (j)66-74.  (n)  79  -  81. 

(c)  26  -  34.  (g)  38  -  42.  (k)  68  .  72.  (o)  42  -  38. 

(d)29-31.  .  (h)  47 -'53.  (1)75-65.  (p)  95  -  75. 

2.  Find  the  value  of  the  following : 

(a)  712-192.  (c)  1462-542.  (e)  12152  -  152. 

(b)  1462-462.  (d)  3122-2882.  (f)  21462-102. 

252.  Different  ways  of  carrying  out  the  same  calculations. 
The   preceding   problems  show  that  the  formula  a2  —  ft2 
=  (a  —  J)  (a  +  J)  provides  us  with  a  method  of  making 
calculations    easier.     In  fact,   the   expressions   which   are 
linked   by   the    equality    sign    in    a2  —  b2  =  (a  —  £)  (a  +  &) 
simply  represent  two  different  ways  of  carrying  out  the 
same  calculations,  of  which  the   one   on  the  right  is  by 
far  the  easier. 

253.  Distinction  between  identity  and  equation.  An  equal- 
ity such  as  a2  —  b2  =  (a  —  5)  (a  +  />)   is  called  an  identity. 
It  represents  two  ways  of  making  the  same  calculation. 
The  statement  is  true  for   all  values  of  a   and  I.    The 
pupil  should  not  confuse  the  meaning  of  an  identity  with 
that  of  an  equation.    Thus  a2  —  4  =  (x  —  2)  (x  +  2)  is  true 
for  all  values  of  x,  but  y?  —  4  =  32  is  a  statement  that  is 
true  only  when  x  =  6  or  x  =  —  6 ;  that  is,  it  is  a  state- 
ment  of  equality  in    some  special  situation ;    it  may  be 


POSITIVE  AND  NEGATIVE  NUMBEKS        205 


the  translation  of  an  area  problem,  a  motion  problem,  an 
alloy  problem,  etc.,  but  it  always  represents  some  concrete 
situation,  whereas  x2  —  4  =  (x  —  2)  (x  +  2)  is  an  abstract 
formula  for  calculation  and  is  true  for  all  values  of  x. 

EXERCISES 

1.  Tell  which  of  the  following  are  equations  and  which  are 
identities : 

(a)  4  x*  -  16  =  20. 

(c)  9z2  +  12*  +  4  =  (3z  +  2V.  _   x2-9 

(d)  4 


2.   Solve,  by  factoring,  the  following  equations : 

(a)  ax  +  bx  =  ac  +  be.          (c)  5  a2x  —  4  b*x  =  10  <>•'•  — 

ca      da  20 

(b)c  +  f/  =  _  +  _.  r(d>§__s 

/6N        1  2        ,        ^ 

1  ;  2x-2      3»-3"r4x-4 


6 

,o 
+  2 


*  254.  Calculating  areas.  The  following  exercises  furnish 
applications  of  the  preceding  work  of  this  chapter. 

EXERCISES 

1.  Show  that  the  shaded  area  A  in  Fig.  171  may  be  ex- 
pressed  as    follows:  A  =(S  —  s)(S  +  s),  where    5   is  a  side 
of  the  large  square  and  s  a  side  of  the 

small  square. 

2.  A  carpet  20  ft.  square  is  placed  in  a 
room  25  ft.  square.    The  uncovered  border 
strip  is  to  be  painted.    Find  the  area  of 
the  strip.    Find  the  cost  of  painting  this 
area  at  80  cents  per  square  yard.    Write  a 
formula  to  be  used  in  calculating  the  cost 
of  painting  similar  strips  at  c  cents  per 

yard,  the  carpet  to  be  x  feet  square  and  the  room  r  feet  square. 


206 


GENERAL  MATHEMATICS 


3.   A  metal  plate  is  cut  as  shown  in  Fig.  172.   If 'a  =10  and 
b  =  2,  what  is  the  area  of  the  plate  ?    In  what  two  ways  may 

the  calculating  be  done  ?  What  is  the      ^ a  ^ 

volume  of  metal  if  the  piece  is  |  in. 
thick  ?  What  is  the  weight  if  a  cubic 
inch  of  the  metal  weighs  20  grams  ? 
Write  a  general  formula  for  a  plate 
cut  in  the  form  of  the  figure,  t  inches 
thick  and  weighing  g  grams  per 
square  inch.  Write  the  result  in  a 
form  which  is  easily  calculated. 


- 


FIG.  173 


...  .  .  FIG.  172 

4.  A    design    pattern    is    cut    in 

the  form   shown  in  Fig.  173.    Calculate    the  area.    Make  a 
verbal  problem  illustrating  this  formula. 

5.  We  can  make  an  application  of  our 
knowledge  of  factoring  in  problems  re- 
lated to  circles,  as  will  be  seen  by  solving 
the  following : 

(a)  The  area  of  a  circle  whose  radius 
is  r  is  Trr2.    What  is  the  area  of  a  circle 
whose  radius  is  R? 

(b)  How  can  you  find  the   area  of   the   ring   shaded   in 
Fig.  174?    Indicate  the  area. 

(c  )  Simplify  the  result  of  (b)  by  remov- 
ing the  monomial  factor. 

(d)  What  is   the  area  of   a  running 
track  in  which  R  =  100  and  r  =  90  ? 

(e)  Calculate  the  area  of  the  shaded 
ring  in  Fig.  174  if  R  =  5.5  in.  and  r  =  5 ; 

if  R  =  3.75  and  r  =  0.25. 

FIG.  174 

6.  Allowing  500  Ib.  to  a  cubic  foot,  find 

the  weight  of  a  steel  pipe  20  ft.  long  if  R  =  12  in.  and  r  —  11  in. 
HINT.    Find  a  rule  or  formula  for  the  volume  of  a  cylinder. 


POSITIVE  AND  NEGATIVE  NUMBERS 


207 


3C 


FlG 


7.  Find  the  weight  of  an  iron   rod  6  ft.  long  cast  in  the 

form  shown  in  Fig.  175  if  a  =  2  in.,  b  —  \  iny  and  c  =  \  in. 

HINT.    Allow  500  Ib.  per  cubic  foot. 

255.  Division  of  polynomials 
illustrated  by  arithmetical  num- 
bers. The  process  of  dividing 
one  polynomial  by  another  may 
be  clearly  illustrated  by  a  long- 
division  problem  in  arithmetic ; 

for  example,  we  shall  consider  67,942 -r-  322.  Ordinarily  we 
divide  in  automatic  fashion,  adopting  many  desirable  short 
cuts  which,  though  they  make  our  work  more  efficient, 
nevertheless  obscure  the  meaning. 

In  multiplication  it  was  pointed  out  that  because  of  our  decimal 
system  the  9  in  67,942  does  not  stand  for  9  units,  but  for  900  units 
or  9  •  102  units.  Similarly,  the  7  means  7000,  or  7  •  103,  etc. 

If  we  arrange  dividend  and  divisor  in  the  form  of  polynomials, 
the  division  may  appear  in  either  of  the  following  forms : 


60000  +  7000  +  900  +  40  +  2 
60000  +  4000  +  400 

3000  +  500  +  40 
3000  +  200  +  20 

300  +  20  +  2 
300  +  20  +  2 


300  +  20  +  2 


200  +  10  +  1 


3-102 


10 


6  •  10*  H 

^4-103H 

-4-102 

2-102  +  10  +  1 

3  •  103  - 
3-103- 

h5-102  +  4-10 
h2-102  +  2-10 

3  -102  +  2-10  +  2 
3  -102  +  2  -10  +  2 

The  student  should  study  the  two  preceding  examples 
carefully  in  order  to  be  better  able  to  understand  the  simi- 
larity of  these  with  the  division  of  algebraic  polynomials 
which  we  shall  now  discuss. 


208  GENERAL  MATHEMATICS 

256.  Division  of  algebraic  polynomials.  The  division  of 
algebraic  polynomials  arranged  according  to  either  the 
ascending  or  the  descending  power  of  some  letter  is  similar 
to  the  preceding  division  of  arithmetical  numbers  ;  thus : 

8  .y4  +  2  y3  +  4  y~ 

4  y3  +  5  y2  +  3  y 


<rl 


as  +  a2/; 


a  +  b 


a?-ab 


-  a*b  -  ab* 

+  aft2  +  bs 
+  ab*  +  b* 

257.  Process  in  division.  From  a  study  of  the  preceding 
exercises  we  see  that  in  dividing  one  polynomial  by  another 
we  proceed  as  follows: 

1.  Arrange  both  dividend  and  divisor  according  to  ascend- 
in;!  '»'  descending  powers  of  some  common  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor  and  write  the  result  for  the  first  term  of  the  quotient. 

3.  Multiply  the  entire  divisor  by  the  first  term  of  the  quotient 
and  subtract  the  result  from  the  dividend. 

4.  If  there  is  a  remainder,  consider  it  as  a  new  dividend 
and  proceed  as  before. 

The  student  should  observe  that  the  process  in  division 
furnishes  an  excellent  review  of  the  other  fundamental 
processes,  inasmuch  as  they  are  necessary  in  almost  every 
division  problem.  They  should  therefore  be  mastered  as 
soon  as  possible. 


POSITIVE  AND  NEGATIVE  NUMBERS        209 

258.  Checking  a  division.     We  shall  now  illustrate  the 
method  of  checking  a  division  : 

Divide  xs  —  3  x2y  +  3  xy2  —  ys  by  x  —  y. 

Xs  —  3  x2y  +  3  xy2  —  ys  \x  —  y 
x3  —     x2;/  |  x'2  —  2  xy  -f  y2 

—  2  x2y  +  3  xy1 


xy2  -  y9 
xy2  —  y3 

First  method  of  checking.  Since  the  division  is  exact 
(that  is,  there  is  no  remainder),  multiply  the  divisor  by  the 
quotient.  If  the  product  equals  the  dividend,  the  problem 
checks.  This  may  be  expressed  in  cases  where  there  is  no 

remainder  by  the  formula  q  x  d  =  D,  or  q  —  —  • 

tZ 

How  would  you  check  if  there  were  a  remainder?  If 
the  answer  is  not  obvious,  try  to  check  similar  problems  in 
long  division  in  arithmetic. 

Second  method  of  checking.    Assume  values  for  x  and  y. 

Let  x  =  5  and  y  —  2. 

Substituting  in  the  example, 


and  d  —  3. 

D  27 

Substituting,  the  formula  q  =  —  becomes  —  =  9,  or  9  =  9. 

d  o 

Since  9  (or  q)  =  9  (  or  —  ),  the  problem  checks. 

\      »/ 

259.  Importance  of  a  thorough  drill  in  division.  In  the 
process  of  division  practically  all  the  principles  of  the  last 
two  chapters  are  involved.  Hence  the  following  exercises 
are  important  as  a  means  of  reviewing  the  fundamental 
laws  of  addition,  subtraction,  multiplication,  and  division. 


210  GENERAL  MATHEMATICS 

EXERCISES 
Divide,  and  check  by  either  method : 

1.  (x2-  llx  +  30) -=-(x-  5). 

2.  (//  -  »f  -  4y  +  4)-<y  -  3y  +  2). 

3.  (fts  +  7  «»  +  18  a  +  40)  -r-  (a1  +  2  a  +  8). 

4.  (9-9x  +  8x2-4x8)-i-(3-2.r). 
5. 

6.  (27 x8  -  54x2//  +  36xy2  - 

7.  (27  x8  +  54  afy  +  36  xy*  +  8  y8)  - '(3  x  +  2  y). 

8.  x8--x-7/. 
9. 

10. 

11.  (1&  +  8  m  -  32  m2  +  32  ms  -  15  m4)  -=-  (3  +  4  m  -  5  m2). 

12.  (x8  +  2xy  +  xz  +  yz  +  if)  +  (x  +  y  +  «). 

13.  (14x  +  2x4  +  llx2  +  5x8  -  24)--(2x2  +'3x  -  4). 

14.  (r8  +  65  r  -  15  r2  -  63)  -s-  (r  -  7). 

15.  (25  a  -  20  a2  +  6  a8  -  12)  --  (-  4  a  +  2  a2  +  3). 

16.  (8x  -  4  +  6x4  +  8x8  -  Ilx2)-r-(4x2  +  2x3  -  x  +  2). 

17.  (9  x2//2  -  6  x8?/  +  x4  -  4  ?/4)  ^  (x2  -  3  xy  + 

18.  (25  x4  -  60  x2/  +  36  y4)  -=-  (5  x2  -  6  y2). 

19.  (4  x4  +  12  a-y  +  9  ?/)  H-  (2  x2  +  3  y2). 

20.  (a5 -1) -=-(«-!);  (aa-l)-s-(a-l). 

21.  (a5-7/5)-(a-7/);  (a*-,ft  +  (a-y). 
.    22.  (25m4-49/i4)^-(5m2  +  7w2). 

23.  (25  m4  -  49  w4)  --  (5  m2  -  7  w2). 

24.  (0.027  aW  +  c8)  -h  (0.2  aft  +  c). 

25.  (8  a8  -  Z>8)  --  (4  a2  +  2  ab  +  62). 


POSITIVE  AND  NEGATIVE  NUMBERS        211 

1     2    —  3    x 

260.  Division  by  zero.  The  quotients  -»  -»  -  — »  -,  •  ••, 

etc.  have  no  meaning,  for  a  number  multiplied  by  0 
cannot  give  1,  2,  —  3,  x,  etc.  (see  the  definition  of  division  in 

Art.  243).  The  quotient  -  is  undetermined,  as  every  num- 
ber multiplied  by  0  equals  0.  Therefore  we  shall  assume 
that  in  all  quotients  hereafter  the  divisor  is  not  zero  nor 
equal  to  zero. 

EXERCISES 

1.  The  following  solution  is  one  of  several  that  are  some- 
times given  to  show  that  1  =  2.    Find  the  fallacy. 

Two  numbers  are  given  equal,  as  x  —  y. 

Then  x  -  y  -  0,  Why  ? 

and  2  (a;  -  y)  =  0.  Why  ? 

Then  x  -  y  =  2  (x  -  y).  Why  ? 

Dividing  both  sides  by  x  —  y,         1  =  2. 

2.  Give  a  similar  argument  which  seems  to  show  that  2 
equals  5. 

SUMMARY 

261.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  turning  tendency,  force,  lever  arm, 
multiplication,  division,  factoring,  factors,  prime  number, 
number  containing  a  monomial  factor. 

262.  The  law  of  signs  in  multiplication  was  illustrated 
(1)  geometrically  with  line  segments  and  (2)  by  means 
of  the  balanced  beam. 

263.  The  following  agreements  were  used : 

1.  A  weight  pulling  downward  is  negative;   one  pulling 
upward  is  positive. 

2.  A  force  tending  to  rotate  a  bar  clockwise  is  negative  ; 
counterclockwise,  positive. 


212  GENERAL  MATHEMATICS 

3.  A  lever  arm  to  the  right  of  the  point  where  the  bar  is 
balanced  is  positive ;  to  the  left,  negative. 

The  turning  tendency  (or  force)  acting  upon  a  balanced 
bar  is  equal  to  the  product  of  the  weight  times  the  lever  arm. 

264.  Law   of  signs    in   multiplication:    The  product   of 
two  numbers  having  like  signs  is  positive  ;  the  product  of  two 
numbers  having  unlike  signs  is  negative. 

265.  The  chapter  has  taught  and  geometrically  illus- 
trated the  following  processes  of  multiplication: 

1.  The  multiplication  of  several  monomials. 

2.  The  multiplication  of  a  monomial  by  a  polynomial. 

3.  The  multiplication  of  polynomials  by  polynomials. 

The  order  of  factors  may  be  changed  without  changing 
the  product. 

The  value  of  a  product  is  zero  if  one  of  the  factors 
is  zero. 

266.  Law  of  division :   The  quotient  of  two  numbers  hav- 
ing like  signs  is  positive  ;  the  quotient  of  two  numbers  having 
unlike  signs  is  negative. 

Arithmetical  numbers  may  be  arranged  in  the  form  of 
polynomials  according  to  powers  of  10. 

The  process  of  dividing  one  polynomial  by  another  is  essentially 
the  same  as  the  process  of  dividing  arithmetical  numbers. 
In  all  problems  of  the  text  the  divisor  is  not  zero. 

267.  The   chapter  has   taught  the  following  forms   of 
division : 

1.  The  division  of  a  monomial  by  a  monomial. 

2.  The  division  of  a  polynomial  by  a  monomial. 

3.  The  reduction  of  a  fraction  to  lowest  terms. 

4.  The  division  of  a  polynomial  by  a  polynomial. 


POSITIVE  AND  NEGATIVE  NUMBERS        213 

268.  Division  has  been  illustrated  geometrically.    Two 
methods  for  checking  division  were  taught. 

269.  The  following  types  of  factoring  were  taught: 
Type  I.    Taking  out  a  common  monomial  factor, 

ax  +  bx  -f  ex  =  x  (a  +  b  +  c). 
Type  II.    The  "  cut  and  try  "  method, 

acx2  +  (be  +  ad~)  x+bd  =  (ax  +  6)  (ex  +  d). 


CHAPTER  X 


GRAPHICAL  REPRESENTATION  OF  STATISTICS;  THE 
GRAPH  OF  A  LINEAR  EQUATION 

270.  Facts  presented  in  the  form  of  a  table.  The  follow, 
ing  table  of  facts  shows  in  part  the  recreational  interests 
of  the  boys  and  girls  of  certain  Cleveland  (Ohio)  high 
schools.  Thus,  of  4528  boys,  4075  play  baseball ;  of  3727 
girls,  2608  play  baseball ;  7402  children  out  of  a  total  of 
8255  attend  the  movies  regularly;  and  so  on. 

TABLE  OF  RECREATIONAL  INTERESTS 


BOYS 

GIRLS 

TOTAL 

Number  of  students  

4528 

3727 

8255 

Number  who  play  baseball    

4075 

2608 

6683 

Number  who  play  basketball     
Number  who  play  tennis   

3018 
1811 

1390 
1864 

4408 
3675 

Number  who  belong  to  Camp  Fire  Girls  . 
Number  who  wrestle  .... 

1358 

621 

621 
1358 

Number  who  attend  movies  

4010 

3392 

7402 

Number  who  attend  movies  daily  .... 

754 

485 

1239 

EXERCISE 

Study  the  preceding  table  until  you  understand  the  mean- 
ing of  the  columns  of  figures. 

271.  Pictograms;  graphs.  Tables  made  up  of  columns  of 
figures  are  common  in  newspapers,  magazines,  and  books, 
but  a  table  like  the  preceding  is  not  the  best  device  for 

214 


REPRESENTATION  OF  STATISTICS  215 

expressing  the  meaning  of  an  array  of  facts.  The  ordinary 
mind  cannot  see  the  relations  if  the  list  is  at  all  extended ; 
hence  it  often  happens  that  the  real  meaning  of  a  series  of 
facts  is  lost  in  a  complicated  table.  Newspapers,  magazines, 
trade  journals,  and  books,  realizing  this  fact,  are  beginning 
to  add  to  tables  of  statistics  pictures  which  show  their 
meaning  and  their  relationships  more  clearly  than  can  be 
done  by  columns  of  figures. 

The  significance  of  the  facts  of  the  preceding  table  is 
far  more  vividly  expressed  by  the  pictures  of  Fig.  176. 
Thus  the  pictures  show  that  of  the  high-school  girls  one 
out  of  every  two  (50  per  cent)  plays  tennis ;  two  out  of 
every  dozen  (16|  per  cent)  are  Camp  Fire  Girls ;  of  the 
high-school  boys  six  out  of  every  twenty  (30  per  cent) 
wrestle ;  85  per  cent  of  all  the  elementary-school  and 
high-school  boys  attend  the  movies  regularly;  and  so  on. 

The  pictures  constitute  a  more  powerful  method  of 
teaching  numerical  relations,  because  they  teach  through 
the  eye.  For  this  reason  they  are  called  graphic  pictures, 
pictograms,  or  simply  graphs. 

EXERCISES 

By  means  of  the  pictograms  in  Fig.  176  answer  the  following 
questions : 

1.  What  per  cent  of  the  Cleveland  boys  play  tennis  ?  of 
the  Cleveland  girls  ?    With  which  group  is  tennis  the  more 
popular  ? 

2.  Assuming  that  every  sound-bodied  boy  should  learn  to 
wrestle,  does  your  class  make  a  better  or  a  worse  showing  in 
per  cents  than  the  Cleveland  boys  ? 

3.  Are  a  larger  per  cent  of  the  girls  of  your  class  Camp 
Fire  Girls  than  is  the  case  in  the  Cleveland  high  schools  ? 


High-School  Girls 
High-School  Boys 
High-School  Girls 
High-School  Boys 
High-School  Girls 


Do  not  play  Tennis 
Do  not  play  Tennis 


We  Never  Do 


We  Never  Do. 


We  Never  Do 


We  play  Baseball  I  Never  Do 

o^o  High-School  Girls     ^^^^^^U^AA, 

Belong  to  Camp  Fire  Do  not  belong  to  Camp  Fire 


High-School  Boys 


Wrestle  Do  not  Wrestle 

Elementary-  and  High-School  Boys 


Attend  Movies 


Daily  Nonattendance 
High-School  Boys  \j 

Help! 

High-School  Girls      .,          laU_^W^_ 
S.  0.  S,  Board  of  Education  ! 


Do  not  attend  Movies 


FIG.  176.    SHOWING  HOW  PICTOGRAMS  ARE  CSED  TO  EXPRESS  FACTS 

(Adapted  from  Johnson's  "Education  through  Recreation  ") 

216 


REPRESENTATION  OF  STATISTICS 


217 


4.  Compare  your  class's  swimming  record  with  the  Cleve- 
land record. 

5.  Continue  the  discussion  with  your  classmates  until  it  is 
clear  to  you  just  how  the  figures  of  the  pictograms  represent 
the  facts  of  the  table. 

272.  The  circle  pictogram.    The  circle  is  frequently  used 
to   show   quantitative    relations.      It   shows   two   things: 

(1)  the  relation  of  each  magnitude  to  each  of  the  others ; 

(2)  the  relation  of  each  magnitude  to  the  sum  of  all. 
The  graph  in  Fig.  177  shows  that  in   a  certain  year 

Portland  (Oregon)  spent  30.7$  out  of  every  dollar  on 
its  school  system.  This  is  prac- 
tically j3^  of  all  the  money  spent 
by  the  city  and  about  1^  times 
as  much  as  the  next  highest 
item.  Compare  the  part  spent 
for  education  with  each  of 
the  other  parts  shown  in  the 
figure. 

The  scale  used  in  making 
circle  pictograms  is  based  on 
the  degree.  The  angular  space 
around  the  center  of  the  circle 
(360°)  is  divided  into  parts  so 
as  to  express  the  numerical  re- 
lations ;  for  example,  since  almost  ^  of  the  money  is 
spent  for  education,  an  angle  of  ^  of  360°,  or  108°,  is 
constructed  with  the  protractor  at  the  center  of  the  circle 
and  the  sides  of  the  angle  extended  until  they  intersect 
the  circle.  The  sector  (the  part  of  a  circle  bounded  by 
two  radii  and  an  arc)  formed  shows  the  part  of  the  city's 
money  that  goes  toward  education. 


FIG.  177.   THE  CIRCLE 
PICTOGRAM 

(From  Cubberly's  "  Portland 
Survey  ") 


218  (iKXERAL  MATHEMATICS 

EXERCISES 

1.  In  Fig.  177  how  does  the  amount  paid  as  interest  com- 
pare with  the  amount  paid  for  police  and  fire  protection  ? 

2.  Do  you  think  it  would  have  been  more  profitable  in  the 
long  run  for  Portland  to  pay  cash  for  all  public  improvements  ? 

HINT.  A  definite  answer  to  this  problem  may  be  obtained  if 
several  members  of  the  class  will  solve  Ex.  3  and  report  to  the  class. 

*3.  What  will  it  cost  a  city  to  build  a  $100,000  high-school 
building  if  $20,000  of  its  cost  is  paid  in  cash  and  the  remainder 
paid  by  issuing  4  per  cent  bonds  of  which  $4000  worth  are  to  be 
retired  (paid)  annually  ?  (All  interest  due  to  be  paid  annually.) 
NOTE.  The  problem  must  not  be  interpreted  as  an  argument 
showing  that  bonding  (borrowing)  is  never  justifiable. 

*4.  The  discussion  of  the  method  of  paying  the  expenses 
of  the  United  States  for  the  first  year  of  our  participation  in 
the  European  War  was  sharply  divided  between  two  groups. 
One  group  favored  a  large  amount  of  borrowing  by  the  issu- 
ance of  bonds,  while  the  other  advocated  a  pay-as-you-go 
policy,  that  is,  raising  the  money  by  taxation.  Debate  the 
merits  of  the  two  plans. 

5.  Show  the  following  facts  by  means  of  a  circle  divided 
into  sectors : 

TABLE  SHOWING  DISPOSITION  OF  THE  GROSS  REVENUE  OF 
THE  BELL  TELEPHONE  SYSTEM  FOR  THE  YEAR  1917 
ITEMS  PER  CENT 

Salaries,  wages,  and  incidentals 50 

Taxes .       7 

Surplus 4 

Materials,  rent,  and  traveling  expenses    ....     20 

Interest 7 

Dividends 12 

Though  widely  used,  the  circle  divided  into  sectors  is 
not  a  quite  satisfactory  method  of  showing  the  ratio  of 


REPKESENTATION  OF  STATISTICS 


219 


numbers.  In  fact,  the  objections  are  so  serious  that  the 
method  of  construction  was  given  to  protect  the  student 
against  false  conclusions.  The  method  is  not  inaccurate 
when  the  parts  which  constitute  a  unit  are  shown  by  the 
use  of  one  circle.  It  frequently 
happens,  however,  that  the  com- 
parison is  made  by  circles  differ- 
ing in  size.  In  such  a  case,  since 
the  eye  tends  to  make  the  com- 
parison on  an  area  basis,  the  ratio  FlG- 178-  CIRCLES  DRAWN  ON 

P.I  v  1111  AN  AREA  BASIS  SHOWING  THE 

of  the  two  numbers  should  be  ex-  BB  op  BANK  DEPOSITORS 

pressed  by  the  ratio  of  the  areas 

of  the  two  circles,  and  statistical  authorities  so  recommend. 
In  Fig.  1 78  the  circles  are  drawn  on  an  area  basis,  but  the 
right-hand  circle  appears  less  prominent  than  the  figures 
justify,  thus  causing  the  reader 
to  underestimate  the  ratio.  In 
Fig.  179  the  circles  are  drawn 
on  a  diameter  basis.  The  right- 
hand  circle  appears  more  prom- 
inent than  the  figures  justify, 
thus  causing  the  reader  to  over- 
estimate the  ratio.  This  feature 
is  frequently  utilized  by  those  who  make  dishonest  use 
of  circle  diagrams.  The  conclusion  is  that  a  comparison 
between  circles  differing  in  size  should  be  avoided  alto- 
gether. Better  graphic  methods  will  be  taught.  Space  is 
given  here  to  circle  pictograms  because  of  their  extensive 
use  in  many  fields. 

EXERCISE 

Test  the  accuracy  of  circle  pictograms  which  you  may  find 
in  magazine  articles  and  advertisements.  Discuss  their  value 
with  your  classmates. 


FIG.  179.    CIRCLES  DRAWN  ON 
A  DIAMETER  BASIS 


220 


GENERAL  MATHEMATICS 


1911 


1899 


273.  Area  pictograms.    The  picture  of  the 
two  traveling  men  given  here  is  intended  to 
show  the  increase  in  the  passenger  traffic  of 
the  railroads.    The  two  men  are  compared 
on  the  basis  of  height. 
The    1911    man,    on 
account    of    his    far 
greater     area,     looks 
more    than   2^  times 
as  large  as  the  1899 
man.  The  men  should 
be   compared   on  the 
basis   of  area.1    This 
type    too    should    be 
avoided  because  it  tends  to  deceive  the  ordinary  reader. 


14,591,000    One  Mile        32,837,000    One  Mile 


FIG.  180.   A  POPULAR  TYPE  OF  PICTOGRAM, 
TO  BE  AVOIDED 


1911 


32.837000 

ONE  MILE 


mmmmM 


1899 


14.59I.OOO 
ONE  MILE 


FIG.  181.    A  MORE  ACCURATE  METHOD  OF  PORTRAYING  FACTS 

EXERCISES 

1.  Why  would  it  be  difficult  to  make  a  drawing  on  the 
basis  of  area? 

2.  Do  you  know  any  method  which  could  be  used  to  check 
a  drawing  made  on  the  basis  of  area  ?  (See  Art.  109.) 

274.  Volume  or  block  pictograms.  Cubes,  parallelepipeds, 
and  spheres  are  frequently  used  in  comparing  relative 
volumes ;  for  example,  pictures  of  bales  of  hay  or  cotton 

1  Brinton,  in  his  excellent  text, ' '  Graphic  Methods  for  Presenting  Facts,' ' 
presents  a  chart  (Fig.  181)  drawn  from  the  same  facts  as  that  in  Fig.  180. 
Note  that  the  facts  are  portrayed  much  more  clearly  and  accurately. 


REPRESENTATION  OF  STATISTICS  221 

are  used  to  show  the  output  of  the  states  producing  these 
articles.  The  comparison  should  be  made  on  a  basis  of 
volume,  but  often  there  is  no  way  for  the  reader  to  tell 
on  what  basis  the  drawing  was  constructed,  whether  by 
height,  area,  or  volume.  Certainly  it  would  be  difficult  to 
check  the  statement  made  in  such  a  case. 

275.  Limitations  of  area  and  volume  pictograms.  The 
student  will  need  to  remember  that  in  a  correctly  con- 
structed area  graph  the  quantities  represented  should  vary 
directly  as  the  number  of  square  units  within  the  out- 
lines of  the  figures.  Thus,  in  the  comparison  of  passenger 
service  relative  size  should  not  be  determined  by  the 
relative  heights  of  the  men  but  by  the  number  of  square 
units  within  the  outlines.  Hence  a  rough  method  of 
checking  is  to  transfer  the  pictures  of  the  traveling  men  to 
squaredpaper  by  means  of  tracing  paper  and  compare  the 
number  of  square  millimeters  in  the  area  of  each  with  the 
corresponding  facts  of  the  table.  Similarly,  in  accurate 
volume  or  block  graphs  the  quantities  should  vary  as  the 
number  of  cubic  units. 

Many  who  use  this  form  of  statistical  interpretation 
carelessly  fail  to  observe  these  principles,  and  the  diffi- 
culty of  a  check  makes  this  form  of  graph  a  convenient 
device  for  those  who  would  dishonestly  misrepresent  the 
facts.  The  general  public  is  not  always  able  to  interpret 
the  graphs  correctly  even  if  they  have  been  properly  drawn. 
Because  of  these  limitations  it  is  somewhat  unfortunate 
that  this  type  of  graph  is  so  extensively  used  in  bulletins 
and  current  magazines. 

EXERCISE 

Try  to  obtain  and  present  to  the  class  an  advertisement 
illustrating  the  misuse  of  an  area  or  volume  pictogram. 


222 


GENERAL  MATHEMATICS 


276.  Practice  in  interpreting  the  bar  diagram.  Fig.  182 
shows  one  of  the  suggestions  of  the  Joint  Committee  on 
Standards  for  Graphic  Presentation.  The  diagram,  Fig.  182, 
(a),  based  on  linear  measurement,  is  called  a  bar  diagram. 
We  shall  study  this  topic  further  in  the  next  article.  Review 


Year 

1900 


1914 


FIG 


Tons 

270.588 


555,031  

(a)  (b) 

182.    BAR  DIAGRAMS  SHOW  FACTS  BETTER  THAN  AREA  AND 
VOLUME  PICTOGRAMS 


the  objections  to  the  other  two  diagrams  (the  squares  and 
blocks  shown  in  Fig.  182,  (b)).  Where  it  is  possible  the 
student  should  represent  quantities  by  linear  magnitudes, 
as  representation  by  areas  or  volumes  is  more  likely  to 
be  misinterpreted. 

EXERCISES 

1.  Study  Fig.  183  and  determine  to  what  extent  the  two 
horizontal  bars  are  helpful  in  expressing  the  ratio  of  the  two 
numbers  given. 

2.  Would    the     UXfl  1  Cotton,  $820,320,000 
bars  in   Fig.  183 

be  sufficient  with- 
out   the    illustra-      1HI"    Wheat,  $561,051,000 
tions  at  the   left 
of  the   numbers  ?        ^IG'  ^'    ^  FAIR  DIAGRAM.    (AFTER  BRINTON) 

3.  With  the  aid  of  compasses  check  the  accuracy  of  Fig.  184. 
Note  that  the  figures  are  written  to  the  left  of  the  bars.   In  many 

woe        $4.409,136  ^1  bar  diagrams  the  figures 

^^^^^^^^^^  are  written  to  the  right 

1912        28  soo  139  •  ^^^^m   °^  ^e  bars.    Can  you 

FIG.  184.1  DIAGRAM  SHOWJNG  .N  Ex-    think  of  a  serious  ob.lec- 

PORTS  OF  AUTOMOBILES.   (AFTER  BRINTON)    tion  to  that  method  ? 

1  See  paragraph  7  under  Art.  277 


REPRESENTATION  OF  STATISTICS 


223 


4.  Why  is  there  a  space  left  between  the  bars  for  1906  and 
1911  in  Fig.  184  ?  Do  you  see  any  other  way  to  improve  the 
diagram  ?  (See  Art.  277.) 

5.  Draw  on  the  blackboard  a  figure  similar  to  Fig.  184, 
adding   a  bar  for  the 

year  1917.  (The  sum 
for  this  year  is  about 
900,000,000.) 

6.  Explain  Fig.  185. 

7.  Show     that     the 
bars  of  Fig.  186  reveal 

more  clearly  than  the  following  table  the  rank  of  the  United 
States  in  respect  to  wealth.     These  are  the  1914  estimates. 

United  States $150,000,000,000 


FIG.   185.     DIAGRAM     SHOWING    DEATH 
KATE  FROM  TYPHOID  IN  1912  PER  HUN- 
DRED THOUSAND  POPULATION 


Great  Britain  and  Ireland 

Germany 

France   

Russia 

Austria-Hungary      .     .     . 
Italy 


85,000,000,000 
80,000,000,000 
50,000,000,000 
40,000,000,000 
25,000,000,000 
20,000,000,000 


8.  Show  that  it  would  have  been  as  accurate  and  more  con- 
venient to  draw  the  preceding  diagrams  on  squared  paper. 


75  100  125          150 


UNITED  STATES  150 

GREAT  BRITAIN  or 
AND  IRELAND-  »5 

GERMANY  80 
FRANCE  -  -  60 

!  •••    <;  '       f        j 

i  ;  -   •  1    '      \    ' 

';•>>•'                \    ; 

:       .    U 

RUSSIA  -.40 

'  --  i 

AUSTRIA-HUNGARY.  25 
ITALY  20 

.  .  .;>;  .. 

mm 

FIG.  186.    COMPARATIVE  WEALTH  OF  NATIONS  IN  1914 

9.  The  table  for  the  wealth  of  nations  contains  estimates 
prepared  at  the  beginning  of  the  European  War  (1914-1918). 


-2'24  -     GENERAL  MATHEMATICS 

These  estimates  are  now  far  from  facts.  The  student  should 
attempt  to  get  the  latest  estimates  from  the  "  World  Almanac  " 
and  construct  a  bar  diagram  which  will  present  the  situation  to 
date  and  will  enable  him  to  make  an  interesting  comparison. 

10.  Discuss  bar  diagrams  similar  to  those  given  on  pages 
222-223  which  you  may  find  in  Popular  Mechanics  Magazine, 
Motor,  Popular  Science  Monthly,  and  Industrial  Management. 
For  the  time  being  limit  yourself  to  the  simpler  diagrams. 

277.  How  to  construct  a  bar  diagram.  An  understanding 
of  how  to  construct  bar  diagrams  and  how  to  interpret 
those  he  may  find  in  newspapers  and  magazines  should 
be  a  part  of  the  education  of  every  general  reader,  just  as 
it  is  of  every  engineer,  physician,  statistician,  and  biologist.1 
As  civilization  advances  there  is  being  brought  to  the  atten- 
tion of  the  reading  public  a  constantly  increasing  amount 
of  comparative  figures  of  a  scientific,  technical,  and  statis- 
tical nature.  A  picture  or  a  diagram  which  presents  such 
data  in  a  way  to  save  time  and  also  to  gain  clearness  is 
a  graph.  The  bar  diagram  is  a  widely  used  method  of  con- 
veying statistical  information  graphically.  The  solution  of 
the  introductory  exercises  along  with  the  discussion  of  such 
supplementary  graphs  as  may  have  seemed  profitable  for  the 
class  to  discuss  will  help  the  pupil  to  understand  the  follow- 
ing outline  of  the  method  of  constructing  a  bar  diagram : 

1.  The  bars  should  be  constructed  to  scale.  To  obtain 
a  convenient  unit  first  inspect  the  size  of  the  smallest 
and  the  largest  number  and  then  choose  a  line  segment  to 

1  Neither  pupils  nor  teachers  should  be  misled  by  the  apparent  sim- 
plicity of  this  work.  The  details  are  of  the  greatest  importance.  It  will 
be  helpful  to  obtain  the  reports  of  the  Joint  Committee  on  Standards  for 
Graphic  Presentation.  This  is  a  competent  committee  of  seventeen,  which 
has  expended  considerable  effort  on  these  elementary  phases.  The  pre- 
liminary report  may  be  had  from  the  American  Society  of  Mechanical 
Engineers,  29  W.  39th  St.,  New  York  ;  price,  10  cents. 


KEPRESENTATION  OF  STATISTICS  225 

represent  a  number  such  that  it  will  be  possible  to  draw  an 
accurate  bar  for  the  smallest  number  and  a  bar  not  too  long 
for  the  largest  number.  The  lines  in  Fig.  186  on  page  223 
are  so  constructed  that  the  relation  between  the  lengths 
of  any  two  is  the  same  as  the  relative  size  of  the  quan- 
tities represented.  A  line  segment  1  mm.  long  represents 
$2,500,000,000  of  wealth.  In  the  table  on  page  223  the 
United  States  is-  estimated  as  possessing  three  times  as 
much  wealth  as  France,  and  so  the  line  segments  repre- 
senting the  wealth  of  the  United  States  and  France  are 
respectively  60  mm.  and  20  mm.  long. 

2.  The  scale  and  sufficient  data  should  appear  on  the 
diagram. 

3.  Each  bar  should  be  designated. 

4.  The  bars  should  be  uniform  in  width. 

5.  The  diagram,  should  have  a  title  or  legend. 

6.  Accuracy  is  the  important  characteristic. 

7.  The  space  between  the  bars  should  be  the  same  as  the 
width  of  the  bars,  except  in  a  case  like  Fig.  184,  where  a 
larger  space  indicates  that  the  three  bars  do  -not  represent 
consecutive  years. 

8.  In  general  the  zero  of  the  scale  should  be  shown.  How- 
ever, there  are  exceptions  ;  for  example,  in  graphing  the  tem- 
perature of  a  patient  we  are  particularly  concerned  with  how 
much  above  or  below  the  normal  the  patient's  temperature 
is.    Hence,  in  a  case  like  this  we  should  emphasize  the 
normal  temperature  line. 

EXERCISES 

1.  Present  the  facts  of  the  table  given  on  page  226  by  means 
of  a  bar  diagram,  using  the  scale  1  mm.  =  f  2,000,000.  The  table 
is  arranged  to  show  the  twenty  heaviest  buyers  of  American 
goods,  as  indicated  by  the  value  of  exports  from  the  United 
States  during  the  fiscal  year  1914. 


GE;NEKAL  MATHEMATICS 


AMERICA'S  TWENTY  BEST  CUSTOMERS 
(From  the  report  of  the  Bureau  of  Foreign  and  Domestic  Commerce) 


VALUE  OF 
PURCHASES 

VALUE  OF. 
PURCHASES 

1.  England     .     . 

$548,641,399 

11.  Argentina      .     . 

$45,179,089 

2.  Germany    . 

344,794,276 

12.  Mexico 

38,748,793 

3.  Canada  .     .     . 

344,716,981 

13.  Scotland    .     .     . 

33,950,947 

4.  France  .     .     . 

159,818,924 

14.  Spain    .     .     .     . 

30,387,569 

5.  Netherlands    . 

112,215,673 

15.  Russia  .... 

30,088,643 

6.  Oceania      .     . 

83,568,417 

16.  Brazil  .... 

29,963,914 

7.  Italy      .     .     . 

74,235,012 

17.  China  .... 

24,698,734 

8.  Cuba     .     .     . 

68,884,428 

18.  Austria-Hungary 

22,718,258 

9.  Belgium     .     . 

61,219,894 

19.  Panama   .      .     . 

22,678,234 

10.  Japan    .     .     . 

51,205,520 

20.  Chile    .... 

17,482,392 

*2.  If  possible,  ascertain  the  facts  to  date  (see  "World  Alma- 
nac "),  graph  results  as  in  Ex.  1,  and  compare  the  two  diagrams. 
Account  for  unusual  changes.  Are  new  customers  appearing 
among  the  "  twenty  best "  ?  Have  old  ones  dropped  out  ? 

3.  Present  the  statistics  of  the  following  table  by  means  of  a 
bar  diagram  showing  the  comparative  length  of  rivers.  (Use  the 
scale  1  cm.  =  400 mi. ;  the  lengths  given  in  the  table  are  in  miles.) 


LENGTH 

LENGTH 

Missouri-Mississippi  .     . 

4200 

Volga  

2400 

Amazon  

3800 

Mackenzie 

2300 

Nile    

3766 

Plata   .     .          ... 

2300 

Yangtze  

3400 

St  Lawrence 

2150 

Yenisei    

3300 

Danube 

1725 

Kongo     

3000 

Euphrates 

1700 

Lena  

3000 

Indus 

1700 

Niger  

3000 

Brahmaputra 

1680 

Ob*    

2700 

Ganges 

1500 

Hoangho 

2600 

Mekon^ 

1500 

Amur      

2500 

Rio  Theodoro 

950 

REPRESENTATION  OF  STATISTICS 


227 


4.  Represent  the  statistics  of  the  following  table  by  bar 
diagrams.  The  estimates  of  the  leading  crops  in  the  United 
States  for  the  year  1917  are  here  compared  with  the  revised 
figures  for  the  crops  of  the  preceding  nine  years.  The  pupil 
should  note  that  each  column  is  a  separate  problem. 


REPORT  OF  THE  UNITED  STATES  DEPARTMENT  OF 
AGRICULTURE  FOR  1917 


YEAR 

CORN 

WHEAT 

OATS 

BARLEY 

RYE 

COTTON 

Bushels 

Bushels 

Bushels 

Bushels 

Bushels 

Bales   .  ' 

1917  .... 

3,159,494,000 

650,828,000 

1,587,286,000 

208,975,000 

60,145,000 

10,949,000 

1916  .... 

2,583,241,000 

639,886,000 

1,251,992,000 

180,927,000 

47,383,000 

12,900,000 

1915  .... 

2,994,793,000 

1,025,801,000 

1,549,030,000 

228,851,000 

54,000,050 

12,862,000 

1914  .... 

2,672,804,000 

891,017,000 

1,141,060,000 

194,953,000 

42,778,000 

15,136,000 

1913  .... 

2,446,988,000 

763,380,000 

1,121,768,000 

178,189,000 

41,381,000 

14,552,000 

1912  .... 

3,124,746,000 

730,267,000 

1,418,377,000 

223,824,000 

35,664,000 

14,104,000 

1911  .... 

2,531,488,000 

621,338,000 

922,298,000 

160,240,000 

33,119,000 

16,101,000 

1910  .... 

2,886,260,000 

635,121,000 

1,186,341,000 

173,832,000 

34,897,000 

12,075,000 

1909  .... 

2,552,190,000 

683,350,000 

1,007,129,000 

173,321,000 

29,520,000 

10,513,000 

1908  .... 

2,668,651,000 

664,002,000 

807,156,000 

166,756,000 

31,851,000 

13,817,000 

278.  Bar  diagrams  used  to  show  several  factors.  We 
shall  now  see  how  bar  diagrams  may  be  used  to  show 
several  factors. 

INTRODUCTORY  EXERCISES 

1.  Fig.  187,  on  page  228,  differs  from  those  in  Art.  277 
in  that  it  presents  two  factors.  What  is  the  scale  of  the  dia- 
gram ?  Note  that  the  bars  representing  new  buildings  extend 
from  the  top  to  the  bottom  of  the  black  bar.   Try  to  account  for 
the  heavy  losses  by  fire  in  1904  and  1906.  Why  is  the  bar  so 
short  for  new  buildings  in  1908  ?   (The  values  are  given  in 
millions  of  dollars.)    Criticize  this  diagram  according  to  the 
principles  of  Art.  277. 

2.  Give  the  facts  of  Fig.  187  for  the  twelfth  year;  the 
eighteenth  year. 


228 


GENERAL  MATHEMATICS 


JNew  Building 


Fire  Losses 


1901    1902   1903   1904 


1905 


1906   1907   1908   1909   1910   1911 


FIG.  187.  DIAGRAM  OF  YEARLY  VALUES  OF  NEW  BUILDINGS,  AND  OF  ALL 
BUILDINGS  LOST  BY  FIRE  IN  THE  UNITED  STATES,  1901-1911,  INCLUSIVE 

(Courtesy  of  W.  C.  Brinton) 

3.  Fig.  188  shows  the  business  relations  involved  when  a  city 
bonds  itself  to  buy  some  present  need  or  luxury.  The  parts 
of  a  single  bar  (say  the  tenth)  show  the  following:  (a)  the 
interest  paid  to  date  (the  black  portion)  ;  (b)  the  amount 
of  the  $75,000  still  outstanding  (the  plain  portion) ;  (c)  the 
part  of  the  debt  that  has  been  paid  (the  crosshatched  portion). 

Show  that  a  public  bond  issue  is  not  only  a  debt  but  that 
it  "  conies  dangerously  near  to  a  perpetual  tax," 


REPRESENTATION  OF  STATISTICS 


229 


19  20 


100.000 
90.000 
80,000 
70,000 
60.000 
50.000 
40,000 
30.000 
20,000 
10.000 


0  "- 

FIG.  188.    BAR  DIAGRAM  USED  TO  SHOW  MONEY  TRANSACTIONS 
INVOLVED  IN  PAVING  FOR  A  $75,000  SCHOOL  BUILDING 

(Adapted  from  Ayres's  "  Springfield  Survey  ") 

The  preceding  exercises  show  how  a  bar  diagram  may 
be  used  to  compare  several  factors  of  some  problem 
which  are  more  or  less  related.  If  the  pupil  is  especially 
interested  in  this  side  of  the  subject,  he  may  do  the 
following  exercises.  The  topic  is  not  particularly  impor- 
tant, however,  because  another  method  which  we  shall 
presently  stftdy  is  much  more  efficient. 

EXERCISES 

*1.  Go  to  the  township,  county,  or  city-hall  authorities  and 
find  out  how  one  or  more  of  your  public  improvements  is 
being  paid  for ;  that  is,  find  out  (a)  if  bonds  were  issued ; 

(b)  how  many  dollars'  worth  are  retired  (paid  for)  each  year ; 

(c)  how  much  interest  must  be  paid  each  year.    Construct  a 


230  GEXEKAL  MATHEMATICS 

bar  diagram  similar  to  the  one  reprinted  from  the  Ayres  report, 
showing  what  it  will  ultimately  cost  your  community  to  pay 
for  the  project. 

*2.  A  certain  county  in  Indiana  built  20  mi.  of  macadam- 
ized road  by  issuing  $40,000  worth  of  4  per  cent  nontaxable 
bonds.  Twenty  $100  bonds  were  to  be  retired  each  year.  By 
means  of  a  bar  diagram  show  how  much  it  ultimately  cost  this 
township  to  build  its  turnpike. 

*3.  Ten  years  after  the  turnpike  referred  to  in  Ex.  2  was 
built  it  was  practically  worn  out.  Did  this  township  lend  to 
posterity  or  borrow  from  it  ?  Give  reasons  for  your  answer. 

279.  Cartograms.  Statistical  maps  which  show  quanti- 
ties that  vary  with  different  geographic  regions  are  some- 
times called  cartograms.  The  student  will  doubtless  find 
examples  in  his  geography.  He  should  also  examine  the 
"  Statistical  Atlas  of  the  United  States,"  published  by 
the  Census  Bureau.  Various  colors  and  shades  are  used 
to  help  express  the  meaning. 

When  the  cost  of  color  printing  is  prohibitive  the  same 
ends  may  be  attained  by  Crosshatch  work.  The  student 
should  examine  rainfall  maps  containing  cartograms  and 
which  are  often  printed  in  newspapers. 

A  special  form  of  cartogram  is  the  dotted  map.  If  \ve 
wish  to  show  the  density  of  population  of  a  city,  we  may 
take  a  map  of-the  city  and  place  a  dot  within  a  square 
for  every  fifty  people  living  in  the  square.  The  scale 
should  be  so  chosen  that  the  dots  will  be  fairly  close 
together  in  the  sections  whose  population  is  of  greatest 
density.  Space  is  not  given  here  to  illustrating  this  type, 
but  the  pupil  will  have  no  difficulty  with  the  exercises 
that  follow. 


REPKESENTATION  OF  STATISTICS 


231 


EXERCISES 

1.  Obtain  at  least  five  different  forms  of  pictograms  and 
cartograms    from  newspapers,   magazines,    trade  journals,   or 
government  bulletins.    Explain  very  briefly  what  each  intends 
to  show. 

2.  Discuss  the  merits  or  defects  of  the  graphs  of  Ex.  1. 

280.  Interpreting  (or  reading)  graphic  curves.  The  intro- 
ductory exercises  given  below  will  furnish  the  student  with 
some  practice  in  the  interpretation  of  graphic  curves. 


1915- 


-1916- 


\7_ 


-1917- 


FIG.  189.    SHOWING    RAILWAY-STOCK    FLUCTUATIONS,    BY   MONTHS,    IN 

THE   AVERAGE  PRICE  OF  TWENTY-FIVE   OF  THE  LEADING   STOCKS  ON 

THE  NEW  YORK  STOCK  EXCHANGE 

(Adapted  from  the  New  York  Times) 


INTRODUCTORY  EXERCISES 


1.  Explain  the  curve  in  Fig.  189,  noting  the  highest  price, 
the  lowest  price,  the  cause  of  the  upward  trend  in  1915,  the 
cause  of  the  downward  movement  in  1917,  and  the  cause  of 
the  sharp  break  upward  in  the  closing  days  of  1917. 


232 


GENERAL  MATHEMATICS 


2.  Explain     the     curve    in 
Fig.  190.     Check    the    graph 
for  the    early  years.    Give   a 
reason  for  such  results  as  you 
may  find. 

3.  Fig.  191  is  a  temperature 
chart  of  a  case  of  typhoid  fever, 
(a)  Explain  the   rise   and  fall 
of  the  curve,    (b)  What  is  the 
meaning  of  the  dots  ?  Do  these 
points  mark  the  tops  of  bars  ? 

(c)  What  assumption  does  the 
physician  make  when  he  con- 
nects these  paints  by  a  curve  ? 

(d)  Note  that  this  diagram  does 
not  have  a  zero  scale ;  why  was 
it  omitted?    The  chart  would 
be  improved  if  it  had  an  empha- 


1910  1911  1912  1913  1914  1915  1916  1917  1918 
90 


(£40 


4.3 


FIG.  190.    NUMBER    OF  CARS   OF 
DIFFERENT  TYPES 

(Adapted  from  Motor) 


sized  line  representing  normal  temperature  (98.4°).  Why  ?  Con- 
struct a  line  in  color  in  your  text  for  the  normal-temperature  line. 


/7 


/jajt&t 


V& 


II  IS 


It/7 


107* 
106' 
105 
104* 
103° 
102* 
101" 

100' 

?£ 

"Itf 


YU590  bl 


&S>  1 


Si/eitr-va 


FIG.  191.    A  "TEMPERATrRE  CHART  OF  A  CASE  OF  TYPHOID  FEVER 


KEPKESENTATIOX  OF  STATISTICS 


233 


4.  Explain  the  heavy  line  (temperature)  of  Fig.  192. 

5.  Explain  the  light  line  (wind)  of  Fig.  192. 

6.  If  your  study  of  science  has  familiarized  you  with  the 
term  "  relative  humidity,"  explain  the  dotted  line  of  Fig.  192. 


SPCCSPCSPCSPCCRSSCS  PCPC  RPCRPCRR    CRRRRR  PCPC 


2    4    6    8    10    12    14   16    18    20   22   24    26   28   80 

FIG.  192.   WEATHER  RECORD  IN  ST.  Louis  FOR  OCTOBER,  1917 

The  heavy  lines  indicate  temperature  in  degrees  F.  The  light  lines  indicate 
•wind  in  miles  per  hour.  The  broken  lines  indicate  relative  humidity  in  percen- 
tage from  readings  taken  at  8  A.M.  and  8  P.M.  The  arrows  fly  with  the  pre- 
vailing direction  of  the  wind.  S,  clear ;  PC,  partly  cloudy ;  C,  cloudy ;  R,  rain. 
(From  the  Healing  and  Ventilating  Magazine,  December,  1917) 

The  curves  of  the  preceding  exercises  are  called  graphic 
curves.  The  graphic  curve  is  particularly  useful  in  com- 
paring the  relation  that  exists  between  ttvo  quantities ;  for 
example,  the  relation  between  the  price  of  wheat  and  the 
annual  yield ;  the  relation  between  the  price  of  beef  and 
the  amount  produced  or  exported;  the  relation  between 
office  expenses  and  the  size  of  the  corporation ;  the  tariff 
necessary  in  order  that  an  article  may  be  manufactured  in 
this  country ;  and  so  on. 

281.  How  the  graphic  curve  is  drawn.  The  graphic  curve 
in  Fig.  193  represents  the  growth  of  the  population  in  the 
United  States  from  1790  to  1910,  as  shown  by  the  table 
on  page  234. 

(1)  The  horizontal  line  OM  represents  the  time  line. 
1  mm.  represents  2£  yr. ;  that  is,  10  yr.  is  represented  by 


234 


GENERAL  MATHEMATICS 


YEAK 

POPULATION 

YEAR 

POPULATION 

1790  .  .  . 

3,929,214 

I860  

31  443  321 

1800  

5,308,483 

1870  

38  558  371 

1810  

7  239  881 

1880 

50  155  783 

1820  

9,638  453 

1890  

62  947  714 

1830  

12,860,702 

1900  

75  994  575 

1840  

1  7  063  363 

1910 

91  972  266 

I860  

23,191,876 

two  small  spaces.  (2)  The  vertical  scale  represents  the 
population  in  millions.  Two  small  spaces  represent  ten 
million.  Therefore  a  bar  about  1.6  mm.  long  is  placed  on  the 
1790  line,  and  a  second  bar  a  little  over  2  mm.  long  is  placed 
on  the  1800  line. 
Similarly,  bars 
were  placed  on 
the  other  lines. 
(3)The  upper  end 
points  of  the  ver- 
tical segments 
(bars)  are  joined 
by  a  curve.  In 
so  far  as  the 
bars  are  con- 
cerned the  fig- 
ure does  not 
differ  from  an 
ordinary  bar  diagram.  We  may  assume,  however,  that  in- 
crease in  population  between  any  two  periods  was  gradual; 
for  example,  we  may  estimate  that  in  1795  the  popula- 
tion was  reasonably  near  some  number  halfway  between 
3,900,000  and  5,300,000 ;  that  is,  about  4,600,000.  Simi- 
larly, we  may  estimate  the  population  in  the  year  1793. 


upr 

so 

y 

Jn 

1 

/ 

I 

/ 

f 

/ 

CO 

/ 

rn 

J/ 

x 

x* 

X 

- 

^ 

x 

^ 

X 

X 

X 

—  *• 

•^ 

r^ 

-—  •  ' 

„ 

r 

1 

0- 

-J 

O- 

| 

s 

J 

_c 

S- 

u 

-0 

5 
I- 

J 

R- 

j 

n- 

0 
-0 

0 
0- 

-j 

:\ 

§. 

-35-. 

FIG.  193.   THE  GRAPHIC  CURVE 


REPRESENTATION  OF  STATISTICS 


235 


This  assumption  leads  us  to  draw  the  smooth  curve  which 
enables  us  to  estimate  the  change  in  population  without 
knowing  the  exact  length  of  the  bars.  By  means  of  the 
curve  predict  what  the  population  will  be  in  1920.  In  what 
way  will  the  accuracy  of  your  prediction  be  affected  by  the 
European  War? 

EXERCISES 

1.  The  following  table  shows  the  total  monthly  sales  of  a 
bookstore  through  a  period  of  two  years. 


January      .... 
February    .... 

1917 
$2125 
2237 

1918 

$2329 
2416 

July     
August    

1917 

$2271 
2231 

1918 
$2380 
2350 

March     

2460 

2479 

September  .... 

2542 

2620 

April       

2521 

2590 

October  ... 

2725 

2831 

May    . 

2486 

2580 

November  .... 

2345 

2540 

June  

2393 

2482 

December   . 

2825 

3129 

Draw  a  graphic  curve  for  each  year  on  the  same  sheet  of 
graphic  paper.  Draw  the  two  curves  with  different-colored 
ink  or  else  use  a  dotted  line  for  one  and  an  unbroken  line  for 
the  other.  Explain  the  curves. 

2.  On  a  winter  day  the  thermometer  was  read  at  9A.M.  and 
every  hour  afterward  until  9  P.  ai.    The  hourly  readings  were 
-  5°,  0°,  -  2°,  -  8°,  -  10°,  -  10°,  -  5°,  0°,  -  5°,  -  4°,  -  2°, 
—  3°,  —  7°.    Draw  the  temperature  graph. 

3.  Using  a  Convenient  scale  and  calling  the  vertical  lines 
age  -lines,  graph  these  average  heights  of  boys  and  girls : 


AGE 

BOYS 

GIRLS 

AGE 

BOYS 

GIRLS 

2yr. 

1.6ft. 

1.6ft. 

12  yr. 

4.8ft. 

4.5ft. 

4 

2.6 

2.6 

14 

5.2 

4.8 

6 

3.0 

3.0 

16 

5.5 

5.2 

8 

3.5 

3.5 

18 

5.6 

5.3 

10 

4.0 

3.9 

20 

5.7 

5.4 

GENERAL  MATHEMATICS 


At  what  age  do  boys  grow  most  rapidly  ?  At  what  age  do 
girls  grow  most  rapidly  ?  Is  it  reasonable  to  assume  that  the 
average  height  of  a  boy  nineteen  years  old  is  5.65  ft.? 

4.  The  standings  of  the  champion  batters  from  1900-1907, 
inclusive,  are  here  given  for  the  National  and  American 
leagues. 

The  National  League : 
0.384     0.382     0.367     0.355     0.349     0.377     0.339     0.350 


The  American  League : 
0.387     0.422     0.376     0.355     0.381 


0.329      0.358     0.350 


Graph  the  data  for  each  league  to  a  convenient  scale,  both 
on  the  same  sheet.    Tell  what  the  lines  show. 

5.  Draw  a  temperature  chart  of  a  patient,  the  data  for 
which  are  given  below  (see  Fig.  191). 


Hour    .    .    . 

IJl'.-M. 

7 

8 

9 

10 

11 

12 

1  A.M. 

2 

3 

4 

3 

Temperature 

100.5' 

101° 

1015° 

103.2" 

102.5° 

101.4° 

101.3' 

101.3° 

101.2° 

101.2° 

101° 

100.7° 

6.  If  possible,  get  a  copy  of  a  temperature  curve  such  as  is 
commonly  kept  in  hospitals  and  explain  the  graph  to  the  class. 
The  class  will  profit  more  by  your  discussion  if  the  curve 
presents  the  data  for  a  long  period. 

7.  Be  on  the  lookout  for  graphic  curves  which  convey  infor- 
mation of  general  interest  to  your  class.    In  nearly  every  news- 
paper you  will  find  tables  of  statistics  which  may  be  plotted  to 
advantage.    Glance  occasionally  through  the  "Statistical  Ab- 
stract of  the  United  States  "  or  the  "  Statistical  Atlas  of  the 
United  States"   (published  by  the  Bureau  of   Foreign   and 
Domestic  Commerce),  Popular  Mechanics  Magazine,  Popular 
Science  Monthly,  Scientific  American,  and  so  on,  with  the  pur- 
pose of  finding  interesting  graphs.    If  a  lack  of  time  prevents 
class  discussion,  post  these  graphs  on  bulletin  boards. 


REPRESENTATION  OF  STATISTICS 


237 


282.  Continuous  and  discrete  series.  We  may  represent 
a  continuous  change  in  wealth,  in  population,  in  the  growth 
of  boys  and  girls,  etc.  by  a  smooth  curve.  Thus,  if  we  read 
four  reports  of  deposits  made  in  a  country  bank  as  $20,000 
on  January  1,  $25,000  on  April  1,  $18,000  on  June  1, 
and  $19,000  on  September  1,  we  assume  that  there  was 
a  gradual  increase  of  deposits  from  January  1  to  April  1, 
a  rather  rigorous  withdrawal  from  April  1  to  June  1,  and  a 
slow  rally  from  June  1  to  September  1.  This  is  precisely 
the  way  a  physician  treats  the  temperature  of  a  patient, 
even  though  he  may  take  the  temperature  but  twice  per  day. 
However,  the  data  of  every  table  cannot  be  considered  as 
continuous  between  the  limits.  This  fact  is  clearly  illus- 
trated by  the  following  table  of  Fourth  of  July  accidents : 


YEAR 

KILLED 

INJURED 

TOTAL 

1909  ....      ... 

215 

5092 

5307 

1910  

131 

2792 

2923 

1911  

57 

1546 

1603 

1912  

41 

945 

986 

1913  

32 

1163 

1195 

1914  

40 

1506 

1546 

If  we  were  to  draw  a  continuous  curve,  it  would  not 
state  the  facts.  Though  a  few  Fourth  of  July  accidents  may 
occur  on  the  third  or  on  the  fifth  of  July,  we  are  certain 
that  a  continuous  curve  would  not  represent  the  facts  for 
the  rest  of  the  year.  Such  a  collection  of  items  is  said  to 
be  a  discrete,  or  broken,  series.  A  record  of  wages  paid  in 
a  factory  is  likely  to  be  a  discrete  series,  for  the  wages  are 
usually  (except  in  piecework)  a  certain  number  of  dollars 
per  week,  the  fractional  parts*  being  seldom  less  than  10  <£. 
We  should  find  very  few  men  getting  odd  sums,  say,  $18.02 
per  week.  Hence  there  would  be  many  gaps  in  the  series. 


238  GENERAL  MATHEMATICS 

283.  Statistics  as  a  science  defined.    We  have  now  pro- 
gressed far  enough  for,  the  student  to  understand  that  the 
term  "  statistics  "  refers  to  a  large  mass  of  facts,  or  data, 
which  bear  upon  some  human  problem.    One  of  the  chief 
uses  of  statistics  as  a  science  is  to  render  the  meaning  of 
masses  of  figures  clear  and  comprehensible  at  a  glance. 
Statistics  gives  us  a  bird's-eye  view  of  a  situation  involv- 
ing a  complex  array  of  numerous  instances  in  such  a  way 
that  we  get  a  picture  which  centers  our  attention  on  a 
few  significant  relations.    Such  a  view  shows  how  one  fac- 
tor in  a  complicated  social  or  economic  problem  influences 
another ;  in  short,  it  enables  us  to  understand  the  relation 
between  variable  (changing)  quantities. 

284.  The  uses  of  statistics.    Statistical  studies  do  not 
exist  merely  to  satisfy  idle  curiosity.    They  are  necessary 
in  the  solution  of  the  most  weighty  social,  governmental,  and 
economic  problems.    Do  certain  social  conditions  make  for 
increase  in  crime  and  poverty  ?  The  sociologist  determines 
statistically  the  relations  bearing  on  the  question.    Are  cer- 
tain criminal  acts  due  to  heredity  ?    The  biologist  presents 
statistical  data.    Is  tuberculosis  increasing  or  decreasing? 
Under  what  conditions  does  it  increase  ?   Reliable  statistics 
presented  by  the  medical  world  guide  our  public  hygienic 
policies.    Further  possibilities  of  statistical  studies  in  the 
medical  world  are  suggested  by  the  recent  work  of  Dr.  Alexis 
Carrel.    The  work  of  Dr.  Carrel  has  been  widely  discussed. 
Though  authorities  disagree  concerning  some  of  the  details, 
all  will  probably  agree  that  the  mathematical  attack  on  the 
problem  of  war  surgery  is  a  distinct  scientific  advance. 

What  insurance  rates  ought  we  to  pay?  Statistical 
investigations  have  determined  laws  for  the  expectation  of 
life  under  given  conditions  which  for  practical  purposes 
are  as  accurate  as  the  formula  for  the  area  of  a  square. 


REPRESENTATION  OF  STATISTICS  239 

The  business  world  at  times  trembles  under  the  threat 
of  gigantic  strikes  that  would  paralyze  all  business.  Are 
the  demands  of  the  men  unreasonable  ?  Are  the  corpora- 
tions earning  undue  dividends  ?  The  public  does  not 
know  and  will  not  know  until  a  scientific  group  of  citizens 
present  reliable  statistics  of  earnings  and  expenditures. 

There  is  now  in  existence  in  Washington  a  nonpar- 
tisan  tariff  commission  which  consists  of  five  members  ap- 
pointed by  the  president,  which  collects  statistics  and 
makes  recommendations  to  Congress  from  time  to  time. 
It  is  now  thought  that  this  commission  may  tend  to  do 
away  with  the.  old  haphazard  methods  of  handling  tariff 
questions. 

How  rapidly  and  with  what  degree  of  accuracy  should 
a  fourth-grade  pupil  be  able  to  add  a  certain  column  of 
figures  ?  The  educator  is  able  to  present  an  answer  based 
on  tests  of  more  than  100,000  fourth-grade  children  for 
that  particular  problem. 

Because  of  the  numerous  trained  enumerators  which 
they  employed  to  cover  the  world's  output,  the  large  spec- 
ulators on  the  Chicago  Board  of  Trade  knew  with  absolute 
certainty  for  days  in  advance  of  the  record-breaking  jump 
in  the  price  of  wheat  in  August,  1916. 

We  might  continue  indefinitely  to  present  evidence 
showing  that  the  intelligent  reader  in  any  field  will  profit 
by  a  knowledge  of  the  elementary  principles  of  statistical 
methods. 

285.  Frequency  table ;  class  limits ;  class  interval.  In 
the  investigation  of  a  problem  it  is  necessary  that  the 
data  be  tabulated  in  some  systematic  fashion  that  will 
enable  us  to  grasp  the  problem.  Suppose  we  measured 
the  length  of  220  ears  of  corn.  Let  us  say  the  smallest 
ear  measures  between  5  in.  and  6  in.,  the  longest  between 


240 


GENERAL  MATHEMATICS 


12  in.  and  13  in.  We  could  then  group  the  ears  by  inches, 
throwing  them  into  eight  groups,  and  tabulate  the  results 
somewhat  as  follows: 

FREQUENCY  TABLE  SHOWING  LENGTH  OF  EARS  OF  CORN 


LENGTH  IN  INCHES 

NUMBER 
OF  EARS 

LENGTH  IN  INCHES 

NUMBER 
OF  EARS 

6-5.99     

2 

9-  9.99      

74 

6-6.99     

4 

10-10.99      ..... 

47 

7-7  99     

20 

11-11.99      ... 

21 

8-8.99     

48 

12-12.99      

4 

70 


CO 


Such  an  arrangement  of  data  is  called  a  frequency  table. 
The  ears  of  corn  have  been  di-       niiiimiiiiiiiiiiiimmiiimiimni 
vided  into  classes.    The  table 
should    be    read    as    follows : 
rt  There  are  two  ears  measuring 
somewhere  between  5  in.  and 
6  in.,  four  between    6  in.   and 
Tin.,"  etc.   The  boundary  lines 
are  known  as  class  limits,  and 
the  distance  between  the  two 
limits  of  any  class  is  designated    40 
as  a  class  interval.    The   class 
interval  in  the  preceding  case 
is  1  in.    Class  intervals  should 
always  be  equal.       « 

The  facts  of  the  table  are 
shown  by  the  graph  in  Fig.  194.  10 
This  graph  is  the  same  as  the 
bar  diagram  (Fig.  186)  which 
we  have  drawn,  with  the  excep- 
tion that  in  this  case  the  bars 
cover  the  scale  intervals. 


so 


20 


7      8      9     10    11    12   13 


FIG.  194.  DISTRIBUTION  OF  FRE- 
QUENCY POLYGON  OF  220  EARS 
OF  CORN 


REPRESENTATION  OF  STATISTICS 


241 


EXERCISES 

1.  What  is  your  guess  as  to  the  shape  of  the  graph  if  the 
corn  were  classified  into  half-inch  groups  ?  into  fourth-inch 
groups  ? 

2.  Are  there  many  very  long  ears  of  corn  ?  many  very  short 
ears  ?    How  do  you  tell  ? 

3.  Into  what  class  interval  does  the  largest  group  fall  ? 

4.  The  following  table   of   frequency   shows   the   lengths 
of  113  leaves  picked  from   a  tree  purely  at  random.    (See 
King's  « Elements  of  Statistical  Method,"  p.  102.) 


LENGTH  OF  LEAF  IN 
CENTIMETERS 

NUMBER  OF 
LEAVES  IN 
EACH  GROUP 

LENGTH  OF  LEAF  ix 
CENTIMETERS 

NUMBER  OF 
LEAVES  IN 
EACH  GROUP 

3-3.99  .... 

4 

8-8.99  .... 

6 

4-4.99  .... 

5 

9-9.99  .... 

4 

5-5.99  .... 

13 

10-10.99  .  .  . 

3 

6-6.99  .... 

56 

11-11.99  .  .  . 

2 

7-7.99  .... 

19 

12-12.99  .  .  . 

1 

The  table  should  be  read  as  follows  :  "  There  are  four  leaves 
between  3  cm.  and  4  cm.  long,  five  between  4  cm.  and  5  cm.,"  etc. 
Graph  as  in  Fig.  194  the  data  of  this  table. 

5.  The  frequency  table  on  page  242  shows  the  weights  of 
1000  twelve-year-old  boys,  1000  thirteen-year-old  boys,  and 
1000  fourteen-year-old  boys.    (The  weights  are  taken   from 
Roberts's  "Manual  of  Anthropometry"  and  include  9  Ib.  of 
clothing  in  each  case.    Figures  are  reduced  to  the  thousand 
basis.) 

6.  Study  each  of  the  groups  for  Exs.  4  and  5  as  to 

(a)  Where  the  smallest  classes  are  found. 

(b)  Where  the  largest  class  is  located. 

(c)  The  gradual  rise  and  fall  of  the  figures. 


( i  KNERAL  M  ATHEMATICS 


FREQUENCY  TABLE   OF  WEIGHTS   SHOWING   1000  TWELVE- 
YEAR-OLD  BOYS,  1000  THIRTEEN-YEAR-OLD  BOYS,  AND  1000 
FOURTEEN-YEAR-OLD  BOYS 


WEIGHTS  IN 
POUNDS 

NUMBER  OF 

12-YEAR-OLD  BOYS 

NUMBER  OF 
13-YEAR-OLD  BOYS 

NUMBER  OF 
14-YEAR-OLD  BOYS 

49-56 

4 

0 

0 

56-63 

24 

6 

0 

63-70 

118 

38 

'3 

70-77 

233 

100 

38 

77-84 

273 

225 

41 

84-91 

221 

256 

130 

91-98 

79 

187 

228 

98-105 

36 

112 

^47 

105-112 

12 

43 

118 

112-119 

0 

17 

82 

119-126 

0 

16 

29 

126-133 

0 

0 

12 

133-140 

0 

0 

18 

NOTE.  The  fact  that  the  third  column  in  the  table  lacks  54  boys  of 
totaling  1000  is  because  54  boys  in  this  group  weighed  over  140  pounds. 
The  49-56  means  from  49  up  to  but  not  including  56'.  Wherever  the 
last  number  of  a  class  is  the  same  as  the  first  number  of  the  next  class, 
the  first  class  includes  up  to  that  point,  but  does  not  include  that  point. 

Construct  the  graph  (similar  to  Fig.  194)  for  each  of  the 
groups  of  the  table  above. 

7.  The  following  test  on  the  ability  to  use  the  four  funda- 
mental laws  in  solving  simple  equations  was  given  to  115  first- 
year  high-school  students,  the  time  given  for  the  test  being 
fifteen  minutes. 

DIRECTIONS  TO  PUPIL 

Find  the  value  of  the  unknown  numbers  in  each  of  the  following 
equations.  Do  not  check  your  results.  Work  the  problems  in  order 
if  possible.  If  you  find  one  too  difficult,  do  not  waste  too  much  time 
on  it,  but  pass  on  to  the  next.  Be  sure  that  it  is  too  difficult,  however, 
before  you  pass  on.  Do  not  omit  any  problem  which  you  can  solve. 


243 


1.  x  +  3  =  7. 

2.  2  #=4. 

3.  2  fc  +  7  =  17. 

4.  z  -  2  =  3. 

5.  2  x  -  4  =  6. 

«.£  =  , 

7    5  y  _  15 


THE  TEST 

13.  16  y  +  2  y  -  18  y  +  2  ij  =  22. 

14.  7x  +  2  =  3z  +  10. 


. 
o 

10.  3.c  +  4|  =9. 

17.  5.3  y  +  0.34  =  2.99. 

18.  0.5x-3  =  1.5. 

19.  3  x  -9^  =  17.5 

20.  7  y  +  20  -  3  y  =  60  +  4  y  +  40  -  8  //. 

21. 


66      18      3 


11.  +  1  =  6. 
4 

12.  |-4=10. 


24. 


25  a; 


The  results  of  the  test  are  given  by  the  following  table 
of  frequency.    The  student  should  study  it  carefully. 


NUMBER  OF 

EXAMPLES 

NUMBER  OF 
STUDENTS 
ATTEMPTING 

NUMBER  OF 
STUDENTS 
SUCCESSFUL 

NUMBER  OF 
EXAMPLES 

NUMBER  OF 
STUDENTS 
ATTEMPTING 

NUMBER  OF 
STUDENTS 
SUCCESSFUL 

0 

0 

0 

13 

1 

11 

1 

0 

1 

14 

4 

15 

2 

0 

1 

15- 

4 

8 

3 

0 

0 

16 

10 

10 

4 

0 

0 

17 

13 

3 

5 

0 

0 

18 

13 

7 

6 

0 

3 

19 

8 

1 

7 

0     • 

4 

20 

22 

2 

8 

1 

5 

21 

13 

0 

9    » 

0 

9 

22 

18 

0 

10 

0 

14 

23 

6 

1 

11 

0 

13 

24 

1 

0 

12 

1 

7 

244  GENERAL  MATHEMATICS 

Explanation.  The  table  consists  of  two  parts,  of  which  the  first 
part  is  the  first,  second,  fourth,  and  fifth  columns,  which  should  be 
read,  "  Of  the  one  hundred  and  fifteen  students  taking  the  test  one 
tried  but  8  examples,  one  attempted  12,  one  attempted  13,  four 
attempted  14,"  etc.  The  second  part  consists  of  the  first,  third, 
fourth,  and  sixth  columns  and  should  be  read,  "  Of  the  one  hundred 
and  fifteen  students  taking  the  test  one  solved  only  1  problem  cor- 
rectly, one  solved  only  2  correctly,  three  solved  6  correctly,  four 
solved  7  correctly,"  etc. 

8.  Construct  a  graph  showing  the  facts  of  the  table  given 
on  page  243  for  the  number  of  students  attempting. 

9.  Under  the  directions  of  your  instructor  take  the  test 
in  Ex.  7. 

10.  Ask  your  instructor  to  give  you  a  frequency  table  show- 
ing the  number  of  attempts  and  successes  in  the  test  taken  by 
your  class  and  determine  how  the  work  done  by  your  class 
compares  in  speed  and  accuracy  with  that  done  by  the  one 
hundred  and  fifteen  students  in  the  test  described  in  Ex.  7. 

286.  Measure  of  central  tendencies ;  the  arithmetic 
average.  A  frequency  table  and  a  frequency  graph  help 
us  to  understand  a  mass  of  facts  because  they  show  us  the 
distribution  of  the  items,  so  that  we  see  where  the  largest 
groups  and  the  smallest  groups  fall.  The  graph  shows  us 
the  general  trend  of  the  facts.  The  large  groups  assume 
importance.  We  need  terms  to  describe  the  central  tend- 
ency. Often  the  word  "  average "  is  used  to  meet  this 
need.  Such  a  term  is  helpful  in  making  a  mass  of  facts 
clear.  Thus,  a  group  of  farmers  could  not  possibly  learn 
much  about  a  field  of  corn  if  we  read  a  list  to  them  show- 
ing the  length  of  every  ear  in  a  field.  But  they  would 
get  some  idea  of  what  yield  to  expect  if  told  that  the 
average  length  of  an  ear  is  91  in.  They  could  certainly 
give  a  fair  estimate  of  the  yield  if  in  addition  we  told 


REPRESENTATION  OF  STATISTICS  245 

them  that  on  tha  average  a  row  in  a  square  40-acre  field 
grew  620  stalks.  We  shall  presently  learn  that  the  word 
"  average,"  as  commonly  used,  is  not  correct.  The  phrase 
"  arithmetic  average  "  means  the  quotient  obtained  by  divid- 
ing the  sum  of  all  the  items  by  the  number  of  items.  Thus, 
to  find  the  average  mark  obtained  by  your  class  on  a  test 
we  need  to  add  the  marks  of  all  the  students  and  divide 
by  the  number  of  students  in  the  class.  If  two  or  more 
students  obtain  the  same  mark  (say  70),  we  can  shorten 
the  first  step  of  the  process  by  multiplying  the  mark  by  the 
number  of  times  it  occurs  instead  of  adding  70  five  times. 
This  means  that  in  a  frequency  table  a  student  must  re- 
member to  multiply  each  item  by  its  frequency  before  adding. 
When  the  size  of  the  item  is  only  approximately  known, 
the  mid-point  of  the  class  interval  is  taken  to  represent 
the  size  of  each  item  therein.  To  illustrate,  suppose  that 
we  should  try  to  find  the  average  number  of  problems 
attempted  in  the  simple-equation  test.  We  shall  suppose 
that  three  students  report  that  they  attempted  6  problems. 
This  does  not  really  mean  that  all  three  exactly  completed 
6  problems  when  time  was  called.  In  all  probability  one 
had  made  a  slight  start  on  number  7,  the  second  was  about 
in  the  middle  of  number  7,  and  the  third  had  almost 
completed  7.  Of  course  the  number  of  students  is  too 
small  to  make  this  certain,  but  if  we  should  take  a  larger 
number  of  students  (say  thirteen),  in  air  probability  there 
would  be  as  many  who  were  more  than  half  through  with 
the  seventh  problem  as  there  would  be  students  less  than 
half  through.  Hence,  to  find  the  average  we  say  that  the 
thirteen  students  attempted  6i  and  not  6,  as  they  reported. 
To  illustrate: 

Find  the  average  number  of  equations  attempted  by  a  class 
on  the  simple-equation  test  if  two  students  report  5  problems 


24G  GENERAL  MATHEMATK  S 

attempted,  four  report  6,  five  report  7,  three  report  8,  and  two 
report  9. 

Solution.  2x5£  =  11 

4  x  6£  =  26 

5  x  7 1 =  37.:. 
:5  x  8i  =  2').-, 
2x9*  =  19 

16  119 

110  -=-16  =7.4. 

Therefore  the  average  number  of  problems  attempted  by  the 
class  is  7.4. 

The  point  is  that  the  series  of  facts  in  the  table  is  not 
a  discrete  series,  as  one  would  at  first  be  inclined  to  think, 
but  a  continuous  series. 

EXERCISES 

*1.  Calculate  the  average  number  of  equations  attempted  by 
your  class  in  the  simple-equation  test  (Art.  285). 

2.  Using  the  table  of  Ex.  5,  Art.  285,  find  the  average 
weight  of  the  twelve-year-old  boys ;  of  the  thirteen-year-old 
boys ;  of  the  fourteen-year-old  boys. 

3.  Find  the  average  length  of  the   113  leaves   in  Ex.  4, 
Art.  285. 

4.  Find  the  average  length  of  the  220  ears  of  corn  of  the 
first  table  in  Art.  285. 

5.  Find  the  average  number  of  Fourth  of  July  accidents 
for  the  six  years  of  the  table  of  Art.  282. 

6.  Compare  the  averages  of  the  champion  batters  for  the 
last  eight  years  (Ex.  4,  Art.  281). 

7.  At   Minneapolis    the    7  A.M.    temperature    readings    for 
the  ten  days  beginning  February  1  were  as  follows :  —  5,  —  3, 
—  5,  —  3,  —7,  —  9,  —  8,  —2,0,  —  6.   Find  the  average  7  A.M. 
temperature  reading  for  the  period. 


REPRESENTATION  OF  STATISTICS  247 

8.  Find  the  average  of  the  following  temperatures:  8 A.M., 
-6°;    9  A.M.,   -5°;    10A.M.,   -2°;    11  A.M.,   -1°;    12  M., 
-2°;  IP.M.,  -4°;  2  P.M.,  -6°;    3  P.M.,  -7°;  4  P.M.,  -7°; 
5  P.  M.,  -  5° ;  6  P.  M.,  -  2° ;  7  P.  M.,  -  1°. 

9.  Find  the  average  church  contributions  according  to  the 
following  frequency  table. 

TABLE  OF  CHURCH  CONTRIBUTIONS 


INDIVIDUAL 
CONTRIBUTIONS 

NTMBER  OF 
CONTRIBUTORS 

INDIVIDUAL 
CONTRIBUTIONS 

NUMBER  OF 
CONTRIBUTORS 

No  contribution  . 
1  cent  

2 

23 

10  cents     .    .    . 
25  cents 

13 

9 

5  cents 

42 

$50    

1 

287.  Disadvantages  of  the  arithmetic  average.  Some  of 
the  preceding  exercises  suggest  that  there  are  certain 
objections  to  the  arithmetic  average.  For  example,  it 
means  little  to  say  that  the  average  church  contribution 
in  Ex.  9,  Art.  286,  is  62  cents.  People  ordinarily  use 
the  word  "  average "  thinking  it  means  the  most  usual 
occurrence ;  that  is,  the  common  thing.  As  a  matter  of  fact 
nobody  gave  62  cents,  and  only  one  person  gave  as  much 
as  that.  The  objection  to  the  arithmetic  average  is  that  it 
gives  too  much  emphasis  to  the  extreme  items.  To  illustrate 
more  fully:  A  boy  who  has  just  finished  an  elementary 
surveying  course  learns  that  the  average  weekly  wage  of  a 
railway-surveying  group  is  $23.  This  is  very  encduraging 
until  an  analysis  shows  him  that  the  chief  engineer  gets  $55 
a  week ;  his  assistant,  |30 ;  and  all  others  but  $15.  To  say 
that  the  average  weekly  earning  of  ten  men  working  in  an 
insurance  office  is  $30  a  week  may  be  misleading,  for  one 
man  may  be  a  $5000-a-year  man,  in  which  case  the  usual 
salary  is  much  lower  than  $30  per  week.  Other  objections 


248  GENERAL  MATHEMATICS 

to  the  arithmetic  average  are  the  following:  (a)  it  cannot 
be  located  either  in  a  frequency  table  or  in  a  frequency 
graph ;  (b)  it  cannot  be  accurately  determined  when  the 
extreme  items  are  missing;  (c)  it  is  likely  to  fall  where 
no  item  exists  (for  example,  a  sociologist  may  discover 
that  the  average-size  family  in  a  given  community  has  4.39 
members,  though  such  a  number  is  evidently  impossible). 
For  these  reasons  it  is  desirable  to  have  some  other 
measure  of  tht  central  tendency  of  a  group. 

288.  Central  tendency ;  the  mode.  One  of  the  most  use- 
ful measures  is  the  mode.  It  may  be  denned  as  the  scale 
interval  that  has  the  most  frequent  item,  or  we  may  say  it 
is  the  place  where  the  longest  bar  of  a  bar  diagram  is 
drawn.  The  term  describes  the  most  usual  occurrence,  or 
the  common  thing.  The  popular  use  of  the  term  "  average  " 
approximates  the  meaning  of  the  word.  When  we  hear  of 
the  average  high-school  boy  he  is  supposed  to  represent 
a  type — one  who  receives  exactly  the  most  common  mark 
of  his  classes,  is  of  the  most  common  athletic  ability, 
spends  the  most  common  amount  of  time  in  study,  shows 
the  most  common  amount  of  school  spirit,  wastes  the  most 
common  amount  of  time,  is  of  the  most  common  age,  etc. 
It  is  obvious  that  no  such  high-school  boy  can  be  found. 
Though  a  boy  may  possess  some  of  these  attributes  he  is 
certain  to  differ  from  the  common  type  in  others. 

The  word  "  average  "  is  thus  incorrectly  used  for  "  mode," 
which  means  the  common  type.  Thus  the  mode  in  the 
church-contribution  table  (Ex.  9,  Art.  286)  is  five  cents. 
More  people  in  this  church  gave  a  nickel  than  any  other 
coin.  The  mode  in  the  frequency  'table  for  the  simple- 
equation  test  for  attempts  (Ex.  7,  Art.  285)  is  twenty 
examples.  In  the  test  more  students  were  at  this  point 
when  time  was  called  than  at  any  other  point. 


REPRESENTATION  OF  STATISTICS  249 

EXERCISES 

1.  Find  the  mode  for  your  class  in  the  frequency  table  for 
successes  in  the  simple-equation  test  of  Ex.  7,  Art.  285. 

2.  From  the  table  of  Ex.  5,  Art.  285,  find  the  mode  for  the 
weight  of  the  twelve-year-old  boys ;   of  the  thirteen-year-old 
boys ;    of  the  fourteen-year-old  boys. 

NOTE.    The  student  is  expected  merely  to  glance  at  the  tables 
to  see  at  what  scale  interval  the  greatest  number  of  items  are  found. 

3.  From  the  table  of  Ex.  4,  Art.  285,  find  the  mode  for  the 
113  leaves. 

4.  From  the  first  table  of  Art.  285  find  the  mode  for  the 
220  ears  of  corn. 

5.  From  the  table  in  Art.  282  find  the  mode  for  the  Fourth 
of  July  accidents. 

289.  Advantages  and  disadvantages  of  the  mode.    The 

advantages  of  the  mode  may  be  summarized  as  follows: 

(a)  the    mode    eliminates    the    extremes.     In  the   results 
of  an  examination  the  mode  is  not  affected  by  the  occa- 
sional hundred  or  zero  marks.    The  salary  of  the  super- 
intendent of  a  division  of  a  shop  does  not  affect  the  mode ; 

(b)  to  the   ordinary  mind  the  mode    means   more    than 
does  an  average.  It  means  more  to  say  that  the  modal  size 
of  classes  in  a  high  school  is  15  than  to  say  that  the  average 
size  is  17.24,  first,  because  there  is  not  a  single  class  that 
actually  has  the  latter  number ;  and,  second,  because  a  few 
large  freshman  classes  in  city  high  schools  may  tend  to 
increase  considerably  the  average  number.   In  making  laws 
we  shall  do  the  greatest  good  to  the  greatest  number  if  we 
keep  the  mode  in  mind  and  not  the  average.   Unfortunately 
street   cars    are   built   for    the    average    number    carried, 


250  GENERAL  MATHEMATICS 

not  for  the  modal  number ;  hence  the  "  strap  hanger." 
The  manufacturer  of  ready-made  clothing  fits  the  modal 
man,  not  the  average  man.  The  spirit  of  a  community's 
charity  fund  is  far  more  evident  in  the  mode  than  in  the 
average. 

A  disadvantage  of  the  mode  is  that  there  are  a  large 
number  of  frequency  tables  to  which  it  cannot  be  easily 
applied.  In  such  cases  we  have  an  irregular  group  with  no 
particular  type  standing  out,  and  the  mode  is  difficult  to 
find,  as  will  be  illustrated  presently. 

290.  Central  tendency ;  the  median.  If  a  number  of  ob- 
jects are  measured  with  reference  to  some  trait,  or  attribute, 
and  then  ranked  accordingly,  they  are  said  to  be  arrayed. 
Suppose  that  your  instructor  gives  an  examination  which 
really  tests  mathematical  ability,  and  that  after  the  results 
are  announced  the  students  stand  in  line,  taking  the 
position  corresponding  to  their  marks  on  the  examination ; 
that  is,  the  student  with  least  mathematical  ability  at  the 
foot  of  the  class,  the  one  next  in  ability  next  to  the  foot, 
etc.  The  class  is  then  arrayed.  If  any  group  of  objects  is 
arrayed,  the  middle  one  is  known  as  the  median  item.  If 
your  class  had  twenty-three  pupils  standing  in  the  order 
of  their  ability,  the  twelfth  pupil  from  the  foot  or  the  head 
of  the  class  is  the  student  whose  mark  is  the  median  mark. 
There  are  just  as  many  below  as  above  him  in  ability.  The 
median  is  another  measure  of  the  central  tendency  of  a 
group.  If  there  is  an  even  number  of  items,  the  median 
is  said  to  exist  halfway  between  the  two  middle  items. 
Thus,  if  your  class  had  twenty-two  pupils,  the  mark  half- 
way between  that  of  the  eleventh  and  that  of  the  twelfth 
student  from  either  end  would  be  called,  the  median  mark. 
The  meaning  is  further  illustrated  by  the  exercises  given 
on  pages  251  and  252. 


KEPKESENTATION  OF  STATISTICS 


251 


1.   Find   the  median    wage  in  the   following   table   of  the 
weekly  wage  of  the  workers  in  a  retail  millinery  shop.1 


WEEKLY  WAGE 

NCMBER  OF 
WORKERS 

. 
WEEKLY  WAGE 

NUMBER  OF 
WORKERS 

$4-5    . 

g 

$11-12     

0 

o-O     

18 

12-13     

5 

(5-7 

16 

13-14              .     .  '  . 

0 

7-8    

12 

14-lo     

6 

8-9    

7 

15-16     

18 

9-10 

8 

16-17         .... 

5 

1(H11  

5 

The  table  above  shows  the  wages  of  one  hundred  girls. 
We  are  asked  to  find  a  weekly  wage  so  that  we  shall  be 
able  to  say  that  one  half  the  girls  in  this  shop  receive  less 
than  this  sum  and  one  half  receive  more  than  this  sum; 
that  is,  we  want  a  measure  of  the  group. 

In  the  first  place,  the  student  should  notice  that  the 
arithmetic  average  $10.05  seems  too  high  to  be  represent- 
ative, for  there  are  too  many  girls  working  for  smaller 
sums.  In  the  second  place,  the  mode  is  unsatisfactory. 
The  wage  $15  to  $16  seems  to  be  a  mode,  but  there  are 
more  girls  working  for  about  $5,  $6,  or  $7 ;  hence  neither 
the  arithmetic  average  nor  the  mode  has  very  much  mean- 
ing, so  we  proceed  to  locate  the  median. 

There  are  one  hundred  girls  in  the  shop  ;  hence  we  must 
find  a  wage  halfway  between  that  of  the  fiftieth  and  that 
of  the  fifty-first  girl  from  the  lowest  wage.  Adding  the 
number  of  the  first  four  groups  of  girls  (3  +  15  +  16+12) 
gives  us  forty-six  girls  and  takes  us  to  the  8-dollar  wage. 
We  need  to  count  four  more  of  the  next  seven,  who  are 

1  For  actual  facts  see  "Dressmaking  and  Millinery,"  in  "The  Cleve- 
land Survey,"  page  63.  The  table  was  adapted  to  meet  the  purposes  of 
the  exercise. 


252  GENERAL  MATHEMATICS 

getting  between  #8  and  $9.  The  table  assumes  that  the 
series  is  continuous  (piecework  ?) ;  hence  the  next  seven 
are  distributed  at  equal  distances  between  $8  and  $9.  We 
may  think  of  the  seven  girls  as  being  distributed  graphi- 
cally, as  shown  in  Fig.  195. 

The  graph  makes  clear  the  assumption  that  the  first  girl 
(the  forty -seventh)  earns  a  sum  which  is  between  §8  and 
$8| ;  we  assume  that  the  wage  is  at  the  mid-point  of  this  in- 
terval, or  $8-Jj.  Similarly,  the  second  girl  (the  forty-eighth) 
earns  a  sum  between  $8|  and 
$8f  and  we  assume  the  wage 
to  be  at  the  mid-point  of  this 
interval,  or  |8^.  In  like  man- 
ner  the  wage  of  the  forty-  F  195 

ninth  girl  is  $8T5¥,  the  fiftieth 

girl  |8T7j,  and  the  fifty-first  girl  $8T9?.  Midway  between 
the  mid-points  of  the  fiftieth  and  the  fifty-first  wage  is 
halfway  between  $8T7¥  and  $8-^,  or  $8|.  Hence  the 
median  is  $8  plus  $|,  or  $84,  for  this  wage  is  halfway 
between  the  wage  of  the  fiftieth  girl  and  that  of  the  fifty- 
first.  The  student  should  study  this  graph  until  this 
point  is  clear.  He  should  note  that  the  average  is  found 
by  calculating,  the  mode  by  inspection,  and  the  median  by 
counting.  Merely  count' along  the  imagined  scale  until  a 
point  is  found  that  divides  the  item  into  two  equal  groups. 
Since  a  wage  problem  usually  involves  a  discrete  series 
(why?),  a  more  practical  illustration  of  the  principle  is 
given  below. 

2.  Find  the  median  for  the  attempts  of  the  one  hundred  and 
fifteen  students  in  the  simple-equation  test  in  Ex.  7,  Art.  285. 

Solution.  We  must  find  the  number  of  equations  attempted  by 
the  fifty-eighth  student  from  either  end,  for  he  will  be  the  middle 
student  of  the  one  hundred  and  fifteen.  Counting  from  the  top  of 


REPRESENTATION  OF  STATISTICS  253 

the  table  (p.  243),  we  get  fifty-five  pupils  who  hive  finished  19 
equations.  We  need  to  count  3  more  to  get  the  fifty-eighth  pupil. 
There  are  twenty-two  more  who  were  somewhere  in  the  twentieth 
equation  when  time  was  called.  If  we  assume,  as  we  did  in  finding 
modes,  that  the  twenty-two  students  are  at  equal  spaces  through- 
out the  twentieth  equation,  then  the  median  is  19  equations  plus 
?\  of  an  equation,  or  just  over  19.1  equations. 

EXERCISES 

1.  State  the  rule  for  finding  the  median,  as  developed  in 
the  two  preceding  exercises. 

2.  In  Ex.  5,  Art.  285,  find  the  median  weight  for  1000 
twelve-year-old    boys;     1000    thirteen-year-eld    boys;     1000 
fourteen-year-old  boys. 

3.  In  Ex.  4,  Art.  285,  find  the  median  leaf  and  its  measure 
in  the  array  of  113  leaves. 

4.  Find  the  median  for  the  220  ears  of  corn  (Art.  285). 

291.  Limitations  of  statistics.  There  is  a  common  saying 
among  nonscientific  people  that  anything  can  be  proved  by 
means  of  statistics.  Experience  lends  conviction  to  the 
homely  saying  "  Figures' do  not  lie,  but  liars  will  figure." 
This  belief  is  due  to  the  fact  that  figures  have  deceived  the 
public  either  by  being  dishonestly  manipulated  or  by  being 
handled  unscientifically.  A  table  dishonestly  manipulated 
or  based  on  unreliable  data  appears  at  first  glance  as  con- 
vincing as  the  work  of  a  trained  scientist.  The  public  does 
not  find  it  possible  to  submit  every  piece  of  evidence  to  a 
critical  study  and  resents  such  deceptions  as  those  referred 
to  above. 

As  a  beginning  the  student  should  determine  (1)  the 
reliability,  and  training  of  those 'who  gathered  the  facts; 
(2)  how  and  when  they  were  gathered  ;  (3)  to  what  extent 


254  GENERAL  MATHEMATICS 

the  statistical  studies  have  been  exposed  to  the  critical 
judgment  of  trained  experts;  (4)  to  what  extent  similar 
studies  show  similar  results. 

292.  Law  of  statistical  regularity.  In  calculating  the 
value  of  the  farm  lands  in  Indiana  it  is  by  no  means 
necessary  to  evaluate  and  tabulate  every  acre  in  the  state. 
To  find  out  the  average  size  of  a  twenty-five-year-old  man 
in  New  York  it  is  not  necessary  .to  measure  and  tabulate 
every  man  in  the  city.  The  "  Statistical  Abstract  of  the 
United  States"  (published  by  the  Bureau  of  Foreign  and 
Domestic  Commerce)  states  the  value  in  dollars  of  hogs, 
sheep,  and  cattle-  produced  in  1918,  but  this  does  not  mean 
that  this  total  is  obtained  by  tabulating  every  individual 
animal.  To  find  out  how  fast  on  an  average  a  twelve- 
year-old  Chicago  boy  can  run  100  yards  we  would  not  need 
to  hold  a  stop  watch  on  every  boy.  In  fact,  a  few  chosen 
in  each  class  in  each  school  building  would  probably  give 
us  an  average  that  would  be  identical  with  an  average 
obtained  from  the  whole  group.  This  is  due  to  a  mathe- 
matical law  of  nature  which  states  that  if  a  reasonable  num- 
ber of  individual  cases  are  chosen  "  at  random  "  from  among 
a  very  large  group,  they  are  almost  sure,  on  the  average,  to  pos- 
sess the  same  characteristics  as  the  larger  groups.  The  phrases 
"  at  random"  and  "reasonable  number"  make  the  law  appear 
somewhat  vague.  King,  in  "Elements  of  Statistical  Method," 
illustrates  the  law  as  follows :  "If  two  persons,  blindfolded, 
were  to  pick,  here  and  there,  three  hundred  walnuts  from 
a  bin  containing  a  million  nute,  the  average  weight  of  the 
nuts  picked  out  by  each  person  would  be  almost  iden- 
tical, even  though  the  nuts  varied  considerably  in  size." 
Gamblers  use  the  principle  just  illustrated  when  they  have 
determined  how  many  times  a  given  event  happens  out  of 
a  given  number  of  possibilities.  They  are  thus  able  to  ply 


REPRESENTATION  OF  STATISTICS  '255 

their  craft  continuously  and  profitably  on  a  small  margin 
in  their  favor.  This  principle  is  the  basis  of  all  insur- 
ance ;  thus  it  is  possible  to  predict  with  a  great  degree 
of  accuracy  how  many  men  of  a  given  age  out  of  a  given 
one  thousand  will,  under  ordinary  conditions,  die  during 
the  next  year.  The  law  of  statistical  regularity  is  very 
extensively  employed  in  the  Census  Bureau.  The  totals 
are  usually  estimates  based  on  careful  study  of  sufficient 
representative  cases. 

However,  the  student  should  be  critical  of  the  phrase 
'"  at  random."  It  is  not  asserted  that  any  small  group 
will  give  the  same  results  as  a  measurement  of  the  whole 
group.  Thus,  if  we  measured  the  height  of  the  first 
four  hundred  men  that  passed  us  as  we  stood  at  the 
corner  of  Randolph  and  State  Streets,  Chicago,  we  could 
not  be  sure  of  getting  an  average  that  would  accurately 
represent  the  city.  Any  number  of  events  might  vitiate  the 
results ;  for  example,  the  Minnesota  football  team  might 
be  passing  by,  or  a  group  of  unusually  small  men  might 
be  returning  from  some  political  or  social  meeting  limited 
to  one  nationality.  The  sampling  should  be  represent- 
ative ;  that  is,  sufficiently  large  and  at  random  (here  and 
there).  The  larger  the  number  of  items,  the  greater  the 
chances  of  getting  a  fair  sample  of  the  larger  group  of 
objects  studied. 

*293.  The  law  of  inertia  of  large  numbers.  This  law 
follows  from  the  law  of  statistical  regularity.  It  asserts 
that  when  a  part  of  a  large  group  differs  so  as  to  show  a 
tendency  in  one  direction,  the  probability  is  that  an  equal 
part  of  the  same  group  has  a  tendency  to  vary  in  the  opposite 
direction;  hence  the  total  change  is  slight. 

294.  Compensating  and  cumulative  errors.  The  preced- 
ing laws  are  also  involved  in  a  discussion  of  errors.  If 


256  GENERAL  MATHEMATICS 

the  pupils  in  your  school  were  to  measure  carefully  the 
length  of  your  instructor's  desk,  the  chances  would  be  that 
as  many  would  give  results  too  large  as  too  short. 

The  estimates  of  a  thousand  observers  of  crop  conditions 
which  are  summarized  or  graphed  in  a  volume  such  as 
the  "  Statistical  Atlas  "  (published  by  the  Department  of 
Commerce)  tend  to  approximate  actual  conditions.  These 
are  illustrations  of  compensating  errors.  "  In  the  long  run 
they  tend  to  make  the  result  lower  as  much  as  higher." 
This  type  of  error  need  not  concern  us  greatly,  provided 
we  have  a  sufficient  number  of  cases. 

However,  we  need  to  be  on  our  guard  against  a  con- 
stant or  cumulative  error.  If  we  use  a  meter  stick  that  is 
too  short,  we  cannot  eliminate  the  error  by  measuring  a 
very  long  line.  A  watch  too  fast  could  not  eventually  be 
a  correct  guide.  A  wholesaler  who  lost  a  little  on  each 
article  sold  could  not  possibly  square  accounts  by  selling 
large  quantities. 

The  value  of  a  mass  of  facts  involving  a  constant  error 
is  seriously  vitiated.  Hence  the  student  should  be  con- 
stantly critical  in  his  effort  to  detect  this  type. 

EXERCISE 

*Draw  as  accurately  as  possible  on  the  blackboard  a  line 
segment  a  certain  number  of  inches  in  length.  Ask  as  many 
as  from  forty  to  fifty  schoolmates,  if  possible,  to  stand  on  a  cer- 
tain spot  and  estimate  the  length  of  the  line.  Find  how  many 
estimated  the  line  too  long ;  how  many  estimated  it  too  short. 

HINT.  The  work  must  be  done  carefully.  Have  each  student 
estimate  four  times ;  that  is,  estimate,  look  away,  estimate,  etc.  Reject 
estimates  of  all  students  who  do  not  comply  seriously  with  your 
request.  How  many  estimated  the  line  too  long  ?  How  many  esti- 
mated it  too  short  ?  Report  the  results  to  your  class. 


KEPRESE^sTATION   OF  STATISTICS 


257 


*295.  Normal  distribution.  The  student  has  observed 
the  regular  rise  and  fall  of  the  numbers  of  the  frequency 
tables  and  the  regular  rise  and  fall  of  the  graphs  which 
express  these  relations.  It  appears  that  the  large  majority 
of  the  items  are  usually  grouped,  and  as  the  distance  from 
this  point  of  grouping  becomes  greater,  the  items  become 
rapidly  fewer  in  number.  If  we  measure  ability  to  solve  a 
set  of  24  simple  equations,  we  shall  discover  that  the  large 
majority  of  any  representative  class  can  attempt  a  little 
more  than  19  in  15  minutes.  Only  a  few,  possibly  none, 
will  try  all  24  problems;  and  only  a  few,  probably  none, 
will  have  attempted 


less   than  6. 
measure    any 


If   we 

single 


FIG.  196.   FREQUENCY  CHART  OF  SCALLOP 

SHELLS  PILED  ACCORDING  TO  THE  NUMBER 

OF  RIBS.    (AFTER  BRINTON) 


human  trait,  we  shall 
discover  the  same 
tendency  in  the  graph 
of  the  results  toward 
a  bell-shaped  curve. 
Whether  we  measure  ability  to  spell,  ability  to  add  a  column 
of  figures,  ability  to  throw  a  baseball,  the  distance  boys  can 
broad-jump,  always  there  are  a  few  very  good  at  it  and 
a  few  very  poor  at  it,  with  the  tendency  of  the  great 
majority  to  possess  but  mediocre  ability  in  a  particular  trait. 
Nature  shows  the  same  tendency.  The  length  of  leaves, 
the  height  of  cornstalks,  the  length  of  ears  of  corn,  etc. 
give  curves  similar  to  the  preceding.  Note  the  piles  of 
scallop  shells  in  Fig.  196.  The  shells  are  sorted  into 
separate  piles  according  to  the  number  of  ribs.  The  piles 
(from  left  to  right)  have  respectively  15,  16,  1.7,  18,  19, 
and  20  ribs.  How  do  you  think  these  piles  would  have 
looked  if  a  greater  number  of  shells  (say  several  hundred) 
had  been  used? 


258 


GENERAL  MATHEMATICS 


The  same  tendency  is  observed  in  economic  affairs. 
Thus,  if  we  measured  the  income  of  the  ordinary  agri- 
cultural community,  we  should  find  out  of  a  thousand 
persons  only  a  few  whose  income  is  less  than  $300  per  year, 
only  a  few,  if  any,  with  an  income  over  $2500,  and  the 
rest  grouped  and 
tapering  between 
these  limits. 

When  the  rise 
and  fall  is  regu- 
lar (that  is,  the 
curve  falls  regu- 
larly on  both  sides 
from  the  mode), 
the  distribution  is 
likely  to  approxi- 
mate what  we  call 
a  normal  distribu- 
tion, and  the  curve 
is  called  a  normal 
distribution  curve. 

A  normal  dis- 
tribution is  illus- 
trated by  the 
table  and  diagram 
(Fig.  197)  given  here,  which  represents  the  heights  from 
actual  measurement  of  four  hundred  and  thirty  Eng- 
lish public-school  boys  from  eleven  to  twelve  years 
of  age.1 

It  will  be  seen  that  the  numbers  conform  to  a  very 
uniform  rule:  the  most  numerous  groups  are  in  the  middle, 
at  53  in.  and  54  in.,  while  the  groups  at  51  in.  and  56  in. 

1  From  Roberts's  "Manual  of  Anthropometry,"  p.  18. 


FIG.  197. 


PHYSICAL  PHENOMENA   ILLUSTRATE 
NORMAL  DISTRIBUTION 


REPRESENTATION  OF  STATISTICS 


259 


are  less  in  number,  those  at  50  in.  and  57  in.  are  still  fewer, 
and  so  on  until  the  extremely  small  numbers  of  the  very 
short  and  very  tall  boys  of  47  in.  and  60  in.  are  reached. 
It  is  shown  that  the  modal,  or  typical,  boy  of  the  class 
and  age  given  is  53.5  in.,  and  since  he  represents  the  most 
numerous  group,  he  forms  the  standard. 

The  curve  would  probably  be  smoother  if  more  boys 
were  measured  or  grouped  into  half -inch  groups.  As  it  is, 
it  approximates  very  nearly  a  normal  distribution. 

Of  course  it  is  not  asserted  that  every  distribution  is 
of  this  type.  There  is  merely  a  tendency  in  chance  and 
in  nature  to  produce  it.  There  are  many  causes  which 
make  distribution  irregular,  as  we  shall  presently  see. 

*296.  Symmetry  of  a  curve.  The  graph  in  Fig.  198 
shows  the  height  of  25,878  American  adult  men  in  inches. 
This  curve,  like  the  one 
of  Art.  295,  is  more  reg- 
ular than  most  curves 
which  we  have  studied. 
It  would  probably  be 
much  smoother  if  the 
class  interval  were  one- 
fourth  inch.  If  we  draw 
a  perpendicular  AK  from 
the  highest  point  of  the  curve,  we  may  think  of  this  as  an 
axis  around  which  the  rectangles  are  built.  The  curve 
to  the  right  of  this  axis  looks  very  much  like  the  part  to 
the  left.  In  this  respect  we  say  the  curve  is  almost 
symmetrical. 

Symmetry  of  figures  may  be  illustrated  by  the  human 
head,  which  is  symmetrical  with  respect  to  a  plane  midway 
between  the  eyes  and  perpendicular  to  the  face ;  thus  the 
left  eye  and  the  left  ear  have  corresponding  parts  to  the 


FIG.  198.    HEIGHT  OF  MEN.    (AFTER 

THORNDIKE,    "  MENTAL   AND  SOCIAL 

MEASUREMENTS,"  p.  98) 


260 


GENERAL  MATHEMATICS 


right  of  this  ti.ris  of  symmetry.  Note  that  the  parts  are 
arrayed  in  reverse  order. 

Other  familiar  illustrations  of  symmetry  are  (1)  the 
hand  and  the  image  obtained  by  holding  the  hand  in  front 
of  a  plane  mirror ;  (2)  words  written  in  ink  and  the  im- 
print of  those  words  on  the  blotting  paper  with  which  they 
are  blotted  :  (3)  our  clothes,  which  are  largely  built  on  the 
principle  of  .symmetry  ;  (4)  the  normal  distribution  curve. 

In  architecture,  in  art,  and  in  higher  mathematics  the 
principle  of  symmetry  is  very  important. 

*297.  Skewness  of  a  curve.  The  term  "skewness"  de- 
notes the  opposite  of  symmetry  and  means  that  the  items 
are  not  symmetrically  distributed.  The  curve  is  not  of  the 
bell-shaped  form.  It  is  higher  either  above  or  below  the 
mode  than  a  sense  of  symmetry  would  have  us  expect. 
To  illustrate:  Snppose  that  the  incomes  of  all  the  people 
living  in  a  certain  community  were  tabulated  as  follows : 


INCO.MK  ix  DOLLARS 

•  XfMBER  OF 

PERSONS 

INCOME  ix  DOLLARS 

N  I'M  HER  OF 
PERSONA 

0-500        .    .    . 

20 

3500-4000      .     .     . 

4 

500-1000      .     .    . 

36 

4000-4500      .     .     . 

3 

1000-1500      .     .     . 

20 

4500-5000      .    .    . 

o 

1500-2000      .    .    . 

12 

5000-5500      .    .    . 

•1 

2000-2500      .    .    . 

6 

5500-6000      .    .    . 

1 

2.300-3000      ... 

5 

6000-6500      ... 

1 

3000-3500     .    .    . 

4 

The  graph  (Fig.  199)  of  this  table  is  not  symmetri- 
cal, but  is  skewed  toward  the  lower  side.  The  meaning  of 
skewness  is  clearly  shown  by  the  graph.  The  graph  no 
longer  presents  the  normal,  symmetrical,  bell-shaped  form ; 
the  base  is  drawn  out  to  a  greater  extent  on  the  one  side 
than  on  the  other. 


REPRESENTATION  OF  STATISTICS 


261 


Distribution  is  often  affected  by  laws  which  are  dis- 
turbing factors  in  the  situation.  Thus,  in  investigating 
the  wages  of  carpenters  we  should  expect  a  few  to  get 
high  wages,  say  90^  per  hour,  and  a  few  very  low,  say  40  <£ 
per  hour,  and  the  rest  to  be  grouped,  according  to  ability, 
between  these  limits.  However,  by  agreement  between 
unions  and  contractors,  carpenters'  wages  are  fixed  in  most 


FIG.  190.    GRAPH  SHOWING  SKE\VNESS  OF  A  CURVE 

cities  at  a  price  somewhere  between  60^  and  85<£.  Hence 
we  should  have  but  one  interval  in  a  distribution  table, 
for  a  particular  city,  say  Minneapolis,  showing  that  all 
carpenters  get  75  <£  per  hour. 

R 

298.  The  graph  of  constant  cost  relations.1  Graphs  may  be 
constructed  and  used  as  "  ready  reckoners  "  for  determining 

1  Teachers  may  find  it  desirable  to  take  up  the  graphing  of  formulas 
from  science  at  this  point;  for  example,  the  graph  of  the  centigrade- 
Fahrenheit  formula.  However,  the  authors  prefer  to  use  a  simpler 
introductory  exercise  here  for  the  sake  of  method  and  to  take  up 
more  purposeful  formulae  in  the  next  chapter. 


262 


GENERAL  MATHEMATICS 


costs  of  different  quantities  of  goods  without  computation. 
This  is  shown  by  the  following  example: 

If  oranges  sell  at  30 $  a  dozen,  the  relation  between  the  num- 
ber of  dozens  and  the  cost  may  be  expressed  by  the  equation 
c  =  30  c/,  where  d  is  the  number  of  dozens  and  c  the  cost 
per  dozen.  If  values  are  given  to  d,  corresponding  values 
may  be  found  for  c,  as  given  in  the  following  table : 


d 

0 

1 

2 

3 

4 

5 

10 

11 

c 

0 

30 

(iO 

90 

120 

150 

300 

330 

On  squared  paper  draw  two  axes,  OX  and  OF,  at  right 
angles.  On  OY  let  a  small  unit  represent  1  doz.,  and  on 
OX  let  a  small  unit  represent  10  $.  Then,  on  the  30  <£  line 


10 


50  100  150  200  250  300 

FIG.  200.   THE  GRAPH  OF  A  COST  FORMULA 


X 


mark  a  point  representing  1  doz.  On  the  20$  line  mark 
a  point  representing  3  doz.  Draw  a  line  through  the  points 
thus  marked.  It  is  seen  that  this  line,  or  graph  (Fig.  200), 
is  a  straight  line. 

By  looking  at  this  price  curve  we  can  get  the  cost  of 
any  number  of  dozens,  even  of  a  fractional  number.  For 
example,  to  find  the  cost  of  6  doz.  observe  the  point  where 


263 

the  horizontal  line  six  small  units  up  meets  the  price 
curve ;  observe  the  point  directly  beneath  this  on  the 
axis  OX',  this  is  eighteen  small  units  from  0  and  hence 
represents  $1.80.  Similarly,  the  cost  of  8£  doz.  is  seen 
to  be  $2.55. 

EXERCISES 

1.  By  means  of  the  graph  in  Fig.  200  determine  the  cost 
of  the  following:  9  doz.;   11  doz.;  2^  doz.;  3^  doz.;   10|doz.; 
5|  doz. ;  3^  doz. 

2.  If  eggs  sell  at  45$  a  dozen,  draw  the  price  graph. 

3.  On  the  price  graph  drawn  for  Ex.  2  find  the  cost  of  4  doz. ; 
3  doz.;  10  doz. ;  3^doz.;  5^  doz. ;  4^  doz. 

4.  Draw  a  price  graph  for  sugar  costing  10|  $  a  pound. 

5.  On  the  graph  drawn  for  Ex.  4  find  the  cost  of  11  lb.; 
31  lb.;  6|lb.;  10  lb. 

6.  Construct  a  graph  which  may  be  used  in  calculating  the 
price  of  potatoes  at  $2.10  per  bushel. 

7.  Use  the  graph  of  Ex.  6  to  find  the  cost  of  3  bu.;  4|-  bu.; 
2  bu.  3  pk. ;  5£  bu. ;  5  bu.  3  pk. 

8.  Since  the  graphs   in  Exs.  1~7  are  straight   lines,   how 
many  of  the  points  would  have  to  be  located  in  each  case  in 
order  to  draw  the  line  ?  Should  these  be  taken  close  together 
or  far  apart,  in  order  to  get  the  position  of  the  price  graph 
more  nearly  accurate  ?  Why  ? 

299.  Graphs  of  linear  equations ;  locus ;  coordinates.  As 
shown  in  the  preceding  sections  the  relation  between  two 
quantities  may  be  expressed  in  three  ways:  (1)  by  an 
ordinary  English  sentence,  (2)  by  an  equation,  or  (3)  by 
a  graph.  The  graph  is  said  to  be  the  graph  of  the  equation. 
A  graph  may  be  constructed  for  each  equation  that  we 


264 


GENERAL  MATHEMATICS 


have  studied  to  date.    The  process  of  drawing  the  graph 
of  an  equation  will  be  given  in  this  article. 

Let  the  equation  be  y  =  2  x  +  3,  which  we  shall  suppose 
is  the  translation  of  some  sentence  which  states  some 
definite  practical  rule;  for  example,  the  cost  of  sending  a 
package  by  parcel  post  into  a  certain  zone  equals  two  cents 
per  ounce  plus  three  cents.  We  want  to  draw  a  graph 
for  the  equation  y  =  2  x  +  3. 


EXERCISES 

1.  What  is  the  value  of  y  in  the  equation  y  =  2x  +  3  when 
x  equals  0  ?  when  x  equals  1  ?  when  x  equals  2  ?  when  x  equals 
3  ?  when  x  equals  —  2  ?  when  x  equals  —  3  ? 

2.  Fill  in  the  following  table  of  values  of  x  and  y  for  the 
equation  y  =  2  x  +  3. 


X 

0 

1 

2 

3 

4 

5 

6 

7 

-  1 

-  2 

-  3 

-  4 

y 

3 

5 

7 

0 

We  are  now  ready  to  transfer  the  data  of  Ex.  2  to 
squared  paper.  The  process  does  not  differ  very  much 
from  our  work  in  frequency  tables  except  that  usually  in 
graphing  equations  we  need  to  consider  both  positive  and 
negative  numbers.  For  the  sake  of  method  we  shall  extend 
the  discussion  to  cover  this  point.  Two  axes,  AA''  and 
)')"'  (Fig.  201),  are  drawn  at  right  angles^and  meet 
at  0.  Corresponding  to  each  set  of  values  of  x  and  y  a 
point  is  located,  the  values  of  x  being  measured  along  or 
parallel  to  XX',  and  the  values  of  y  along  or  parallel  to 
YY'.  -Positive  values  of  x  are  measured  to  the  right  of 
YY'  and  negative  values  to  the  left ;  positive  values  of  y 
are  measured  above  XX'  and  negative  values  below  A'A"'. 


REPRESENTATION  OF  STATISTICS 


265 


m 


I 


For  example,  the  point  A  corresponding  to  x  ^=  1  and  y  =  5 
is  obtained  by  measuring  one  space  to  the  right  and  five 
spaces  upward.  The  point  B  corresponding  to  x  =  —  1  and 
y  =  1  is  obtained  by  measuring  one  space  to  the  left  and 
one  space  upward.  The 
point  C  corresponding 
to  x  =  —  3  and  y  =  —  3 
is  obtained  by  measur- 
ing three  spaces  to  the 
left  and  three  spaces 
downward.  The  x  and 
y  values  of  each  point 
are  called  the  coordi- 
nates of  that  point. 

Continue  finding 
points  which  represent 
corresponding  parts  of 
numbers  in  the  table. 
tt  soon  becomes  ap- 
parent that  all  the  points  seem  to  lie  on  a  straight  line. 
Hence  we  need  to  plot  only  two  points  to  be  able  to 
draw  the  line.  However,  we  are  more  certain  to  discover 
possible  errors  if  we  plot  three  points.  Why? 

,     EXERCISES 

1.  Find  the   values   of  x  and  y   at  the   points  D  and  E 
(Fig.   201)   by  inspection.     Determine   whether  they  satisfy 
the  equation  of  the  graph. 

2.  Select  any  point  in  the  line  and  determine  whether  the 
values  of  x  and  y  at  this  point  satisfy  the  equation. 

3.  Select  any  point  not  on  the  graph,  find  the  values  of  x 
and  ylat  this  point,  and  determine  whether  they  satisfy  the 
equation  of  the  graph. 


FIG.  201.    GRAPH  OF  A  PARCEL  POST 
FORMULA 


260  GENERAL  MATHEMATICS 

4.  Select  any  point  of  the  graph  and  determine  whether  the 
values  of  x  and  y  satisfy  the  equation. 

5.  How  many  points  could  one  find  on  the  line  ? 

The  preceding  exercises  illustrate  the  following  facts : 

(a)  The  coordinates  of  every  point  on  the  line  satisfy  the 
equation. 

(b)  The  coordinates  of  every  point  not  on  the  line  do  not 
satisfy  the  equation. 

These  two  facts  can  be  proved  rigidly  in  advanced 
mathematics,  and  they  enable  us  to  say  that  the  straight 
line  found  is  the  lo<nis  (the  place)  of  all  points  whose  coordi- 
nates satisfy  the  given  equation.  It  is  important  to  observe 
that  the  idea  of  a  locus  involves  two  things,  specified  under 
(a)  and  (b)  above. 

Since  it  appears  that  the  graph  of  an  equation  of  the 
first  degree  having  two  unknowns  is  a  straight  line,  equa- 
tions of  the  first  degree  are  called  linear  equations. 

A  line  may  be  extended  indefinitely  in  either  direction, 
and  there  are  an  indefinitely  large  (infinite)  number  of  points 
upon  a  straight  line.  Since  the  coordinates  of  each  point  on 
the  line  satisfy  the  equation  of  the  line,  there  are  an  infinite 
number  of  solutions  of  a  linear  equation  with  two  unknowns. 
This  fact  is  evident,  also,  because  for  every  value  of  one  of 
the  unknowns  we  can  find  a  corresponding  value  for  the 
other  unknown. 

ORAL  EXERCISES 

1.  What  is  the  location  (locus)  of  all  points   in  a  plane 
which  are  at  a  distance  of  5  ft.  from  a  given  point  P  in  the 
plane?  at  a  distance  of  7%  ft.  fromP?  at  a  distance  of  x  feet,? 

2.  What  is  the  locus  of  all  points  in  space  at  a  distance 
of  10  ft.   from  a  given   point  ?    1  cm.   from   a  given  point  ? 
x  yards  from  a  given  point  ? 


REPRESENTATION  OF  STATISTICS 


267 


3.  What  is  the  locus  of  all  points  in  a  plane  3  in.  distant  from  a 
given  straight  line  in  the  plane  ?  5|  in.  distant  ?  y  inches  distant  ? 

4.  What  is  the  locus  of  all  points  in  space  at  a  distance  of 
4  in.  from  a  given  straight  line  ?  6  cm.  ?  y  feet  ? 

5.  What  is  the  locus  of  all  points  in  a  plane  equally  dis- 
tant from  two  given  parallel  lines  in  the  plane  ? 

*6.  What  is  the  locus  of  all  points  in  space  equally  distant 
from  two  given  parallel  lines  ? 

7.  What  is  the  locus  of  all  points  in  a  plane  6  in.  distant  from 
each  of  two  given  points  in  the  plane  which  are  10  in.  apart  ? 

8.  What  is  the  locus  of  all  points  3  ft.  from  the  ceiling  of 
your  classroom  ? 

*9.  What  is  the  locus  of  all  points  in  a  plane  5  in.  distant 
from  a  line  segment  7  in.  long  in  the  plane  ? 

*10.  What  is  the  locus  of  all  points  in  space  5  in.  distant 
from  a  line  segment  10  ft.  long  ? 

300.  Terms  used  in  graphing  a  linear  equation.  Certain 
terms  used  in  mathematics  ,in  connection  with  graphical 
representation  will  now 
be  given  and  illustrated 
by  Fig.  202.  The  lines 
XX'  and  YY',  drawn  at 
right  angles,  are  called 
axes  (XX1  the  horizontal 
axis  and  YY'  the  verti- 
cal axis).  The  point  0  is 
called  the  origin.  From  P, 
any  point  on  the  squared 


-x 


±2 


paper,  perpendiculars  are 

drawn  to   the   axes:    the     FlG-  202'  I""MRATISG  ™K  TERMS 

USED  IN  PLOTTING  A  POINT 

distance  PM  is  called  the 

ordinate  of  P,  and  the  distance  PN  is  called  the  abscissa  of  P  ; 


2b'8  GENERAL  MATHEMATICS 

together  they  are  called  the  coordinates  of  P.  The  axes  are 
called  coordinate  axes.  The  scale  used  is  indicated  on  the 
axes.  The  distances  on  OX  and  0  Y  are  positive ;  those  on 
OX'  and  on  OY'  negative.  The  abscissa  of  P  is  2  and  the 
ordinate  is  2*-;  the  point  P  is  called  the  point  (%,  2±). 
Notice  that  the  abscissa  is  written  first  and  the  ordinate 
second.  Finding  a  point  on  a  graphic  sheet  which  corre- 
sponds to  a  given  pair  of  coordinates  is  called  plotting 
the  point. 

EXERCISES 

1.  What  is  the  abscissa  of  point  A  ?  B  ?  C  ?  D  ?  E  ?  (Fig.  202). 

2.  What  is  the  ordinate  of  the  point  A?  B?  C?  D?  E :' 
(Fig.  202.) 

3.  Give  the  coordinates  of  points  A,  B,  C,  D,  E  (Fig.  202). 

4.  On  a  sheet  of  graphic  paper  draw  a  set  of  coordinate  axes 
intersecting  near  the  center  of  the  paper,  and  plot  the  following 
points  :  (2,  4),  (5,  2),  (4,  -  2),  (-  3,  4),  (-  3,  -  2),(-  2J,  3J). 

5 .  Compare  the  process  of  plotting  points  with  the  numbering 
of  houses  in  a  city. 

6.  On  a  sheet  of  graphic  paper  locate  the  points  A  (2,  2), 
B(5,  3),  C(2,  7),  and  D(5,.8).    What  kind  of  figure  do  you 
think  is  formed  when  the  points  A,  B,  C,  and  7)are  connected? 
Draw  the  diagonals  of  the  figure,  and  find  the  coordinates  of 
the  point  where  the  diagonals  intersect. 

301.  Summary  of  the  method  for  the  process  of  graphing 
a  linear  equation.  With  Art.  300  in  mind  we  shall  now 
illustrate  and  summarize  the  process  of  graphing  a  linear 
equation. 

Draw  the  graph  of  4  x  —  3  y  =  6. 

(a)  Solve  the  equation  for  either  unknown  in  terms  of  the  other:  thus, 


REPRESENTATION  OF  STATISTICS 


269 


This  throws  the  equation  into   a  form  from  which  the  corre- 
sponding pairs  of  values  are  more  easily  obtained. 


(b)  Let 
Then 
And  let 
Then 


x  —  3,  etc. 


That  is,  build  a  table  of  corresponding  values  as  follows :    (Try  to 
get  at  least  two  pairs  of  integral  numbers.    Why?) 


6  +  3  77 


•(c)  Plot  the  points 

corresponding    to    the 

•      f        t       f.j,         FIG.  203.    GRAPH  OP  THE  LINEAR  EQUATION 
pairs  oj  numbers  oj  the  .         „     _  , 

table  (see  Fig.  203). 

(d)  To  check,  choose  a  point  on  the  line  drawn  and  determine 
whether  its  coordinates  satisfy  the  given  equation  or  plot  a  third 
pair  of  numbers  in  the  table.    This  third  point  also  should  fall  on 
the- line  drawn. 

(e)  The  two  points  plotted  should  not  be  too  near  each  other.    Why  f 


270  GENERAL  MATHEMATICS 

> 

EXERCISES 

Draw  the  graphs  of  the  following  equations,  each  on  a  sep- 
arate sheet  of  squared  paper : 

1.  x  +  y  =  7.  5.  5x-  4y  =  20.  9.  5x-2y  =  -3. 

2.2x  —  y  =  &.  6.  3x  +  5y  =  l5.  10.  x  +  5 y  —  — 12. 

3.  3x-  2y  =  l2.  7.  5x-2y  =  W.  11.  2x  =  3-4»/. 

4.  3x  +  2y=6.  8.  6x-4y  =  3.  12.  3y  =  4  — 8ar. 

HISTORICAL  NOTE.  Statistics  has  attained  the  dignity  of  a  sci- 
ence during  the  last  fifty  years.  Its  growth  goes  hand  in  hand  with 
national  organization.  Even  in  a  crude  tribal  organization  the  ruler 
must  needs  know  something  of  its  wealth  to  determine  the  taxes  or 
tribute  which  may  be  levied.  Our  earliest  statistical  compilations 
(some  time  before  3000  B.C.)  presented  the  population  and  wealth 
of  Egypt  in  order  to  arrange  for  the  construction  of  the  pyramids. 
Many  centuries  later  (about  1400  B.C.)  Rameges  II  took  a  census 
of  all  the  lands  of  Egypt  to  reapportion  his  subjects. 

In  the  Bible  we  read  how  Moses  numbered  the  tribes  of  Israel 
and  of  the  census  of  the  Roman  emperor,  Augustus  Caesar,  in  the 
year  which  marked  the  birth  of  Christ. 

The  Greeks  and  Romans  and  the  feudal  barons  of  the  Middle 
Ages  made  many  enumerations  for  the  purposes  of  apportioning 
land,  levying  taxes,  classifying  the  inhabitants,  and  determining  the 
military  strength.  In  all  cases  except  that  of  the  Romans  some 
special  reason  existed  for  collecting  the  data.  The  Romans  col- 
lected such  data  at  regular  intervals. 

During  the  Mercantile  Age  of  western  Europe  the  feeling  grew 
that  it  was  the  function  of  a  government  to  encourage  the  measures 
aimed  to  secure  a  balance  of  trade.  In  order  to  decide  correctly 
concerning  the  needs  of  commercial  legislation,  more  detailed  infor- 
mation was  necessary  than  had  hitherto  been  gathered.  The  growth 
in  a  centralized  monarchy  further  stimulated  statistical  study.  That 
monarch  was  most  successful  who  could  in  advance  most  accurately 
compare  his  resources  with  his  rivals'. 

In  1575  Philip  II  of  Spain  made  extensive  inquiries  from  the  prel- 
ates concerning  their  districts.  In  1696  Louis  XIV  required  reports 
on  the  conditions  of  the  country  from  each  of  the  general  intendants. 


REPRESENTATION  OF  STATISTICS  271 

Prussia  began  in  modern  times  the  policy  of  making  periodic 
collections  of  statistical  data.  In  1719  Frederick  William  I  began 
requiring  semiannual  reports  as  to  population,  occupations,  real- 
estate  holdings,  taxes,  city  finance,  etc.  Later  these  data  were  col- 
lected every  three  years.  Frederick  the  Great  also  was  a  vigorous 
exponent  of  the  value  of  statistics.  He  enlarged  the  scope  of  statis- 
tics in  general  by  including  nationality,  age,  deaths  and  their  causes, 
conditions  of  agriculture,  trade,  manufactures,  shipping,  in  fact, 
anything  that  might  possibly  contribute  to  national  efficiency. 

A  provision  in  our  constitution  of  1790  initiated  the  decennial 
census.  One  country  after  another  has  adopted  some  form  of  regular 
enumeration,  until,  in  1911,  China  took  her  first  official  census. 

In  recent  times  the  censuses  have  grown  extremely  elaborate. 
In  1900  the  United  States  established  a  permanent  Census  Bureau 
whose  function  it  is  to  study  special  problems  in  the  light  of  the 
data  collected  and  to  publish  the  results  of  this  study.  Most  leading 
nations  also  have  special  bureaus  which  attempt  to  keep  the  sta- 
tistics of  a  nation  u\>  to  date  by  means  of  scientific  estimates.  An 
example  of  such  a  bureau  is  our  National  Bureau  of  Statistics. 
Many  states  have  established  bureaus  to  meet  the  needs  of  the 
state.  Recently  a  movement  has  gained  momentum  to  establish 
municipal  bureaus  to  collect  and  study  the  data  of  the  community 
and  to  instruct  the  public  as  to  the  significant  results  obtained  by 
means  of  elaborate  reports.  An  example  of  this  idea  is  illustrated 
by  the  Survey  Committee  of  the  Cleveland  Foundation. 


SUMMARY 

302.  Chapter  X  has  taught  the  meaning  of  the  following 
words  and  phrases:  pictogram,  cartogram,  bar  diagrams, 
graphic  curve,  frequency  table,  class  interval,  central  tend- 
ency, arithmetic  average,  mode,  median,  normal  distri- 
bution, random  sampling,  compensating  errors,  constant 
or  accumulating  errors,  symmetry,  symmetry  of  a  curve, 
skewness  of  a  curve,  price  graph,  linear  equation,  locus, 
axes,  horizontal  axis,  vertical  axis,  ordinate,  abscissa,  coor- 
dinates, coordinate  axes,  plotting  a  point. 


L'7_'  GKNKKAL   MATHEMATICS 

303.  The  graphic  curve  may  be  used  to  show  the  rela- 
tion between  two  quantities.   Specific  directions  were  given 
showing  how  a  graphic  curve  is  drawn. 

304.  Continuous  and  discrete  series  were  illustrated  and 
explained. 

305.  Statistical  studies  are  necessary  to  solve  our  social, 
governmental,  and    economic    problems.     The    intelligent 
reader   will  profit   by  a  knowledge    of  the    elements    of 
statistical  methods. 

306.  Tabulating  the  facts  bearing  on  a  problem  in  the 
form  of  a  frequency  table  enables  one  to  get  a  grasp  on 
the  problem. 

307.  The  word  "  average  "  as  generally  used  may  mean 
arithmetic  average,  mode,  or  median.    All  are  measures  of 
the  central  tendency  of  a  mass  of   statistical  data.    The 
arithmetic  average  is  found  by  figuring,  the  mode  is  found 
by  inspection,  and  the  median  by  counting. 

308.  The  law  of  statistical  regularity  was  illustrated. 

309.  The  law  of  inertia  of  large,  numbers  was  stated. 

310.  The  graph  of  goods  purchased  at  a  constant  cost 
may  be  used  as  a  "  ready  reckoner." 

311.  The   chapter  has  taught  how   to   plot  points   on 
squared  paper. 

312.  The  graph  of  a  linear  equation  is  a  straight  line, 
The  coordinates  of  every  point  on  the  line  satisfy  the  equa- 
tion, and  the  coordinates  of  every  point  not  on  the  line  do 
not  satisfy  the  equation.    This  illustrates  the  locus  idea. 

313.  The  chapter  has  taught  the  method  of  graphing  a 
linear  equation. 


«       CHAPTER  XI 

GAINING  CONTROL  OF  THE  FORMULA;  GRAPHICAL 
INTERPRETATION  OF  FORMULAS 

314.  The  formula.  The  formula  has  been  defined  as  an 
equation  which  is  an  abbreviated  translation  of  some  prac- 
tical rule  of  procedure.    Thus,  I  =  Prt  is  a  formula  because 
it  is  an  equation  which  is  an  abbreviated  form  of  the  fol- 
lowing practical  rule  for  finding  interest :  To  find  the  interest 
(expressed  in  dollars)  multiply  the  principal  (expressed  in 
dollars)  by  the  product  of  the  rate  (expressed  hi  hundredths) 
and  the  time  (expressed  in  years). 

The  formula  is  applied  extensively  in  shop  work,  engi- 
neering, science,  and,  in  fact,  in  every  field  of  business  and 
industry  where  the  literature  is  at  all  technical.  The  sym- 
bolic form  of  the  rule  of  procedure  is  not  only  more  easily 
understood  than  the  sentence  form  but  is  more  easily  applied 
to  a  particular  problem. 

315.  Applying  the  interest  formula.  A  formula  is  applied 
to  a  problem  when  the  known  facts  of  the  problem  are  sub- 
stituted in  place  of  the  letters  of  the  formula.    A  formula 
may  be  used  when  all  the  letters  except  one  appear  as 
known  facts  in  the  problem.    The  pupil  should  study  the 
following  illustration : 

What  is  the  interest  on  $200  at  5%  for  two  years  ? 

Solution.   Substituting  the  known  facts  in  the  formula, 

7  =  200.^-2. 
Simplifying  the  right  member, 

/  =  $20. 
273 


274  GENERAL  MATHEMATICS 

EXERCISES 

1.  Find  the  interest  on  $425  at  4%  for  2£  yr. 

2.  Find  the  interest  on  $640  at  4£%  for  Syr. 


4.5  9 

HINT.    Substitute  —  ^—  ,  or  —  —  ,  for  r.    Why  ? 

3.  Find  the  interest  on  $820  at  4%  for  2yr.  3  mo.  5  da. 

HINT.  Reduce  2  yr.  3  mo.  5  da.  to  days,  divide  this  result  by  360, 
and  substitute  for  /.  Why? 

316.  Other  types  of  interest  problems  conveniently  solved 
by  special  forms  of  I=Prt.  The  method  of  solving  other 
types  of  interest  problems  is  illustrated  by  the  following 
problem  : 

How  much  money  must  be  invested  at  5%  for  2  yr.  so  as  to 
yield  $180  interest  ? 

NOTE.  This  problem  differs  from  Ex.  3,  Art.  315,  in  that  rate, 
time,  and  interest  are  given  and  the  problem  is  to  find  P  (the  prin- 
cipal). It  may  be  solved  by  substituting  the  three  numbers  given 
for  the  corresponding  three  letters  of  the  formula.  Why  ?  However, 
it  will  be  found  on  trial  to  be  far  more  convenient  if  we  first  solve 
for  P  in  /  =  Prt. 

Solution.    Dividing  both  members  of  the  equation  by  rt,  —  —  P. 

•  rt 

This  may  be  translated  into  the  following  rule  of  arithmetic  : 
To  find  the  principal  divide  the  interest  by  the  product  of  the  prin- 

cipal and  the  rate.    P  =  —  is  only  a  special  form  of  I  =  Prt,  but 

constitutes  complete  directions  for  finding  the  principal  when  the 
other  three  factors  are  given. 

In  the  proposed  problem  we  obtain,  by  substituting, 


Thus  the  principal  is  $1800. 


CONTROL  OF  THE  FORMULA  275 

EXERCISES 

1.  What  principal  must  be  invested  at  4|-%   for  2  yr   to 
yield  $81? 

2.  What  is  the  principal  if  the  rate  is  6%,  the  time  4yr. 
3  da.,  and  the  interest  $120  ? 

3.  What  is  the  rate  if  the  principal  is  $500,  the  time  3  yr., 
and  the  interest  $90  ? 


Here  P,  t,  and  /  are  given  ;  r  is  the  unknown.    Hence  we  solve 
/  =  Prt  for  r. 

Dividing  both  members  by  P  and  then  by  t  or  by  (Pf), 

J_ 

Pt~ 

I 

Substituting  the  known  facts  in  r  =  —  , 

Pt  <<\ 

1)0  6        ™ 

=  -  =  6%. 


500  •  3      100 

4.  Translate  r  =  —  into  a  rule  of  procedure  for  finding 
the  rate. 

5.  What  is  the  rate  if  the  interest  is  $85.50,  the  time 
l|-yr.,  and  the  principal  $950? 

6.  What  is  a  fourth  type  of  interest  problem  ?    Find  a 
formula    most    convenient    for    the    solution    of    such    type 
problems. 

7.  Show  how  to  obtain  this  formula  from  the  form  7  =  Prt. 

8.  Translate  into  a  rule  of  arithmetic.  , 

9.  Into  what  two  parts  can  1500  be  divided  so  that  the 
income  of  one  at  6%   shall  equal  the  income  of  the  other 
at  4%  ? 

10.  How  can  a  man  divide  $1800  so  that  the  income  of  part 
at  4%  shall  be  the  same  as  that  of  the  rest  at  5,%  ? 

11.  A  certain  sum  invested  at  4|%  gave  the  same  interest  in 
2  yr.  as  $4000  gave  in  1^  yr.  at  4^  .    How  large  was  the  sum  ? 


276 


GENERAL  MATHEMATICS 


317.  Solving  a  formula.    The  process  of  deriving  t  =  — 

-LV 

from  /  =  Prt  is  called  solving  the  formula  for  t.  Similarly, 
deriving  the  form  P  —  —  is  called  solving  the  formula 

/  L 

for  P.  The  special  form  obtained  is  not  only  the  most 
convenient  form  for  the  particular  problem,  but  it  may  be 
used  to  solve  the  whole  class  of  problems  to  which  it 
belongs.  The  solving  of  the  formulas  of  this  chapter  are 
of  the  practical  kind  and  will  involve  little  more  than 
the  applications  of  the  axioms  of  Chapter  I. 

318.  Graphical   illustration   of   interest  problems.     The 
relation   between  any  two  of  the  factors   that  appear  in 
an  interest  formula  may  be  represented  graphically. 


„  EXERCISE 

How  does  the  yearly  interest  vary  on  principals  invested 
at  5%  ? 

Substituting  T§5  for  r,  and  1  for  t, 
then  /=  T§(j  P. 

Note  that  this  is  a  linear  equa- 
tion involving  7  and  P  which  may 
be  plotted  by  the  method  of  Art.  301. 
The  table  below  was  used  to  make  the 
graph  in  Fig.  204. 

/  P 


$2.50 

$50.00 

$5.00 

$100.00 

$10.00 

$200.00 

10     15     20     25 
Interest 


Let  one  small  unit  on  the  horizontal 

lines  represent  $1  of  interest,  and  one 

.        . 

large  unit  on  the  vertical  lines  repre-     1N  CALCULATING  INTEREST  ON 
sent  $50  of  principal  invested.  .        PRINCIPALS  INVESTED  AT  5% 


.   0n/<     n 
G.  204.   GRAPH  TO  BE  USED 


'  CONTROL  OF  THE  FORMULA  277 

Use  OX  as  the  line  for  plotting  interests  and  OY  for  plotting  prin- 
cipals. Then  the  point  corresponding  to  ($2.50,  $50)  means  2^  small 
spaces  to  the  right  and  1  large  space  up.  Since  we  know  that  the 
graph  will  be  a  straight  line,  the  line  OR  may  be  safely  drawn  as 
soon  as  tw'o  points  are  plotted. 

EXERCISES 

\.  Look  at  the  graph  in  Fig.  204  and  tell  offhand  what 
interest  you  would  expect  to  collect  at  5%  for  1  yr.  on  $300; 
on  $350 ;  on  $400 ;  on  $60 ;  on 


2.  Determine  by  looking  at  the  graph  in  Fig.  204  how  much 
money  you  would  need  to  invest  at  5%  to  collect  $18  interest 
in  1  yr.;  $20  interest;  $27.50  interest;  $14  interest. 

3.  How  would  you  go  about  finding  the  interest  on  $12.50 
by  means  of  a  graph  ?  on  $2000  ? 

4.  Check  some   of  the  answers  given   by  calculating  the 
interest  by  the  usual  method. 

5.  Graph  the  equation  /  =  T^p  and  use  the  graph  to  cal- 
culate interest  on  sums  lent  at  6%. 

6.  Let  P  =  $100  and  r  =  Tf  7  m  the  formula  1  =  Prt,  thus 
obtaining  I  =  6t.    Graph  7  =  6 1  and  use  the  graph  to  deter- 
mine the  interest  on  $100  at  6%  for  2  yr. ;  for  2|  yr. ;  for  3  yr. ; 
for  4  yr. ;  for  5  yr. ;  for  2  mo. 

7.  If  possible,  report  in  detail  the  methods  used  by  your 
family  banker  to  calculate  interest.    On  what  principles  do  the 
various  "  short  cuts  "  rest  ? 

*319.  Formulas  involving  the  amount.  In  the  exercises 
that  follow  we  shall  study  some  formulas  a  little  more  diffi- 
cult to  solve,  but  they  can  be  understood  if  the  funda- 
mental laws  in  solving  equations  are  carefully  applied. 


278  GENERAL  MATHEMATICS 

EXERCISES 

1.  If  $400  is  invested  at  4%,  what  is  the  amount  at  the 
end  of  1  yr.  ?  of  2  yr.  ?  of  3  yr.  ? 

2.  If  $1200  is  invested  at  3%,  what  is  the  amount  at  the 
end  of  2  yr.  ? 

3.  What  is  the  rule  for  finding  the  amount  when  principal, 
rate,  and  time  are  given  ? 

4.  Using  A  for  the  amount  and  Prt  for  the  interest,  translate 
the  preceding  rule  into  a  formula. 

5.  The  formula  for  the  amount  may  also  be  written  in  the 
form  A  =  P  (1  +  rt~).    Prove. 

6.  Solve  A  =  P  (1  +  rt)  for  P. 

HINT.    Dividing  both  members  of  the  equation  by  the  coefficient 
of  P,  namely,  (1  +  rf),  we  obtain  P  =  —  - 

7.  Translate  into  a  rule  of  arithmetic  the  formula  obtained 
in  Ex.  6. 

8.  Find  the  principal  if  the  rate  is  6%,  the  time  3  yr.,  and 
the  amount  $472. 


9.  Find  the  principal  if  the  rate  is  5%,  the  time  3yr.,  and 
the  amount  $1150. 

10.  Solve  the  equation  .4  =  7*  +  Prt  for  t.    Translate  the 
resulting  formula  into  words. 

11.  Find  the  time  if  the  principal  is  $2500,  the  amount 
$2725,  and  the  rate  3%. 

12.  Solve  the  equation  A  =  P  +  Prt  for  r  and  translate  the 
resulting  formula  into  words. 

13.  Find  the  rate  if  the  principal  is  $1500,  the  amount 
$1740,  and  the  time  4  yr. 

14.  Summarize  the  advantages  of  solving  interest  problems 
by  formulas. 


CONTROL  OF  THE  FORMULA  279 

320.  Evaluating  a  formula.    The  process  of  finding  the 
arithmetical  value  of  the  literal  number  called  for  in  a 
formula  is  called   evaluating  the  formula.    The  foregoing 
exercises  show  that  the  process  consists  of 

1.  Substituting  the  known  numbers  in  the  formula. 

2.  Reducing  the  arithmetical  number  obtained  to  the  simplest 
form. 

NOTE.    A  drill  list  involving  these  processes  is  given  in  Art.  329. 

321 .  Summary  of  the  discussion  of  a  formula.  Cultivating 
and  gaining  control  of  a  formula  means 

1.  Analyzing  an  arithmetical  situation  so  as  'to  see  the 
rule  of  procedure. 

2.  Translating  the  rule  into  a  formula. 

3.  Solving  the  formula  for  any  letter  in  terms  of  all  the 
others. 

4.  Evaluating  the  formula. 

These  steps  will  now  be  illustrated  in  the  solution  of 
motion  problems.  We  shall  then  proceed  to  solve  short 
lists  of  exercises  which  should  develop  power  in  these  steps. 

322.  The  formula  applied  to  motion  problems.   In  solving 
the  following  problems  try  to  observe  the  steps  summarized 
in  Art.  321. 

ORAL  EXERCISES 

1.  If  a  220-yard-dash  man  runs  the  last  50  yd.  in  5  sec.,  at 
what  rate  is  he  finishing  ? 

2.  If  an  automobile  makes  75  mi.  in  2|-hr.,  how  fast  is  it 
being  driven  ? 

3.  Express  the  distance  covered  by  a  train  in  8hr.  at  an 
average  rate  of  20 mi.  per  hour;  of  12^  mi.  per  hour;  of  x  miles 
per  hour ;  of  x  -f  3  mi.  per  hour. 


280  GENERAL  MATHEMATICS 

4.  Express  the  distance  covered  by  a  train  in  t  hours  at  the 
rate  of  /•  miles  per  hour. 

5.  Express  the  time  it  takes  an  automobile  to  go  150  mi.  at  the 
rate  of  10  mi.  per  hour ;  of  15  mi.  per  hour ;  of  20  mi.  per  hour; 
of  m  miles  per  hour ;  of  2  m  miles  per  day ;  of  2  m  +  4  mi.  per  day. 

6.  How  long  does  it  take  to  make  a  trip  of  d  miles  at  the 
rate  of  r  miles  per  hour  ? 

7.  The  rate  of  a  train  is  30  mi.  an  hour.    If  it  leaves  the 
station  at  1  A.M.,  how  far  away  is  it  at  2A.M.;  at  3  A.M.;  at 
1  A.M.:    at  5A.M.;  etc.?    How  far  away  is    it  at  3.15A.M.; 
at  4.30A.M.;  at  6.45A.M.? 

8.  Denoting  the  distance  traveled  by  d,  find  d  when  the  rate 
is  45  mi.  an  hour  and  the  number  of  hours  is  six. 

323.  Distance,  rate,  time.  The  preceding  exercises  show 
that  a  problem  involving  motion  is  concerned  with  distance, 
rate,  and  time.  The  number  of  linear  units  passed  over  by 
a  moving  body  is  called  the  distance,  and  the  number 
of  units  of  distance  traversed  may  be  represented  by  d. 
The  rate  of  uniform  motion,  that  is,  the  number  of  units 
traversed  in  each  unit  of  time,  is  called  the  rate  (or  speed) 
and  is  represented  by  r.  The  time,  t,  is  expressed  in 
minutes,  hours,  days,  etc. 

ORAL  EXERCISES 

1.  Illustrate  by  familiar  experiences  that  distance  equals  the 
rate  multiplied  by  the  time  ;  that  is,  that  d  =  rt. 

2.  Show  how  to  obtain  t  =  —  from  d  =  rt. 

r     • 

3.  Translate  t  =  —  into  a  rule  for  finding  the  time. 

4.  Show  how  to  obtain  r  =  -  from  d  =  rt. 

5.  Translate  r  =  -  into  a  rule  for  finding  the  rate. 


CONTROL  OF  THE  FORMULA  281 

6.  A  motor  cycle  goes  110  mi.  in  5  hr.  and  30  min.  Assum- 
ing the  rate  to  be  uniform,  what  is  the  rate  ?  Which  of  the 
formulas  did  you  use  ? 

7.  Sound  travels  about  1080  ft.  per  second.   If  the  sound  of 
a  stroke  of  lightning  is  heard  2.5  sec.  after  the  flash,  how  far 
away  is  the  stroke  ?  Which  form  of  the  formula  is  used  ? 

8.  How  many  seconds  would  it  take  the  sound  to  reach  the 
ear  if  a  tree  2376  ft.  distant  were  struck  by  lightning  ?  Which 
form  of  the  formula  is  used  ? 

WRITTEN  EXERCISES 

1.  A  motor  boat  starts  10  mi.  behind  a  sailboat  and  runs 
14  mi.  per  hour,  while  the  sailboat  makes  6  mi.  per  hour.    How 
long  will  it  require  the  motor  boat  to  overtake  the  sailboat  ? 

Let  x  be  the  number  of  hours  it  takes  the  motor  boat  to  overtake 
the  sailboat. 

Then,  according  to  the  data  of  the  problem : 

for  the  motor  boat,  for  the  sailboat, 

t  =  x.  t  =  x. 

r  =  14.  r  =  6. 

Hence          d  =  14  x.  Hence         d  —  6  x. 

Since  the  motor  boat  must  go  10  mi.  more  than  the  sailboat,  the 
following  equation  expresses  the  conditions  of  the  problem : 
14  x  =  6  x  +  10. 

Solve  the  equation  to  find  the  value  of  x,  which  turns  out  to 
be  llhr. 

2.  A  and  B  live  22^  mi.  from  each  other.    In  order  to  meet 
A,  B  leaves  home  an  hour  earlier  than  A.   If  A  travels  at  the 
rate  of  4  mi.  an  hour  and  B  at  the  rate  of  3|  mi.  an  hour,  when 
and  where  will  they  meet  ? 

3.  A  northbound  and  a  southbound  train  leave  Chicago  at 
the  same  time,  the  former  running  4  mi.  an  hour  faster  than 
the  latter.    If  at  the  end  of  1^  hr.  the  trains  are  126  mi.  apart, 
find  the  rate  of  each. 


282  GENERAL  MATHEMATICS 

4.  In  running  280  mi.  a  freight  train  whose  rate  is  |  that 
of  an  express  train  takes  3-|-  hr.  longer  than  the  express  train. 
Find  the  rate  of  each. 

5.  An    automobile    runs    10  mi.    an    hour   faster    than   a 
motor  cycle,  and  it  takes  the  automobile  2  hr.  longer  to  run 
150  mi.  than  it  takes  the  motor  cycle  to  run  60  mi.    Find  the 
rate  of  each. 

6.  A  man  rows  downstream  at  the  rate  of  8  mi.  an  hour 
and  returns  at  the  rate  of  5  mi.  an  hour.    How  far  downstream 
can  he  go  and  return  if  he  has  5|  hr.  at  his  disposal  ?   At  what 
rate  does  the  stream  flow  ? 

7.  Chicago  and  Cincinnati  are  about  250  mi.  apart.   Suppose 
that  a  train  starts  from  each  city  toward  the  other,  one  at  the 
rate  of  30  mi.  per  hour  and  the  other  at  the  rate  of  35  mi.  per 
hour.    How  soon  will  they  meet  ? 

8.  A  train  is  traveling  at  the  rate  of  30  mi.  an  hour.    In 
how  many  hours  will  a  second  train  overtake  the  first  if  the 
second  starts  3  hr.  later  than  the  first  and  travels  at  the  rate 
of  35  mi.  an  hour  ? 

*9.  A  and  B  run  a  mile  race.  A  runs  20  ft.  per  second,  and 
B  19^  ft.  per  second.  B  has  a  start  of  32  yd.  In  how  many 
seconds  will  A  overtake  B  ?  Which  will  win  the  race  ? 

*  10.  A  bullet  going  1500  ft.  per  second  is  heard  to  strike 
the  target  3  sec.  after  it  is  fired.  How  far  away  is  the  target  ? 
(Sound  travels  at  the  rate  of  about  1080  ft.  per  second.) 

11.  A  motor  cyclist  rode  85  mi.  in  5  hr.    Part  of  the  distance 
was  on  a  country  road  at  a  speed  of  20  mi.  an  hour  and  the 
rest  within  the  city  limits  at  10  mi.  an  hour.    Find  how  many 
hours  of  his  ride  were  in  the  country. 

12.  Two  boats  149  mi.  apart  approach  each  other,  leaving  at 
the  same  time.    One  goes  10  mi.  per  hour  faster  than  the  other, 
and  they  meet  in  2  hr.    What  is  the  rate  of  each  ? 


CONTROL  OF  THE  FORMULA 


283 


30 


45 


60 


324.  Graphical  illustration  of  a  motion  problem.  Many 
motion  problems  can  be  conveniently  illustrated  graphi- 
cally, as  the  student  will  discover  if  he  solves  the  following 
exercises. 

EXERCISES 

1.  In  the  Indianapolis  races  De  Palma  drove  his  car  at  a 
rate  varying  but  little  from  90  mi.  per  hour.    Draw  a  graph 
showing  the  relation  between  the  distance  and        d         t 
time  of  De  Palma's  performance. 

Substituting  90  in  d  =  tr, 
d  =  9Qt. 

Note  that  d  —  90 1  is  a  linear  equation  which 
may  be  graphed  (see  table  and  Fig.  205).  Ten  small 
units  on  the  vertical  axis  represent  30  mi.;  ten  small  units  on  the 

horizontal  axis  represent  ^  hr. 

• 

2.  Determine  from  the  graph  in  Fig.  205  how  many  miles 

De  Palma  drove  in  2  hr. ;  in  1^  hr. ;  in  1  hr.  24  min. ;  in  40  min. ; 
in   4  min.;    in    2  hr.  12  min. 

3.  Determine  by  the  graph 
in  Fig.  205  how  long  it  took 
De  Palma  to  go  50  mi. ;  40  mi. ; 
GO  mi.;  75  mi.;  140  mi.;  IGOmi.; 
10  mi. 

Obviously  the  preceding  re-  % 
suits  could  be  calculated  either 
by  arithmetic  or  by  the  formula. 
However,  the  graph  has  the  advan- 
tage of  revealing  all  the  results 
in  vivid  fashion. 


i        l       1J 
Time  in  Hours 


FIG.  205.  THE  GRAPH  OF  A  MOTION- 
PROBLEM  FORMULA 


4.  Draw  a  graph  showing 
the  distances  traversed  by  a  passenger  train  running  uniformly 
at  the  rate  of  40  mi.  per  hour  for  the  first  ten  hours  of  its  trip. 
*5.  Find  out,  if  possible,  what  use  railroad  officials  make  of 
graphs  in  arranging  schedules. 


284  GENERAL  MATHEMATICS 

325.  Circular  motion.  Circular  motion  is  of  frequent 
occurrence  in  mechanics.  A  familiar  illustration  is  found 
in  the  movement  of  the  hands  of  a  clock. 

EXERCISES 

1.  At  what  time  between  3  and  4  o'clock  are  the  hands  of 
a  clock  together  ?  h 

Solution.  Let  x  (Fig.  206)  equal  the 
number  of  minutes  after  3  o'clock  when 
the  hands  are  together  ;  that  is,  x  equals 
the  number  of  minute  spaces  over  which 
the  minute  hand  passes  from  3  o'clock 
until  it  overtakes  the  hour  hand. 

x 
Then  —  equals  the  number  of  minute 

FIG.  206.    CLOCK  PROBLEMS 
spaces  passed  over  by  the  hour  hand.    ILLUSTRATE  A  TYPE  OF  ClR. 

Why?  CULAK  MOTION 

Since  the  number  of  minute  spaces 

from  12  to  3  is  15,  and  since  the  whole  is  equal  to  the  sum  of  its 
parts,  it  follows  that 


Whence  x  =  16  ^4T  min. 

Therefore  the  hands  are  together  at  16^  min.  after  3  o'clock. 

2.  At  what  time  between  4  and  5  o'clock  are  the  hands  of 
a  clock  together  ? 

HINT.    Draw  a  figure  similar  to  the  one  for  Ex.  1. 

Notice  that  the  formula  for  a   clock   problem    is    x  =  —  +  m, 

\.£t 

where  m  equals  the  number  of  minute  spaces  the  minute  hand  must 
gain  in  order  to  reach  the  desired  position. 

3.  At  what  time  between  2  and  3  o'clock  are  the  hands  of 
a  clock  15  min.  apart  ? 

HINT.    Draw  a  figure,  think  the  problem  through,  and  then  try 
to  see  how  the  formula  in  Ex.  2  applies. 


CONTROL  OF  THE  FORMULA  285 

4.  At  what  time  between  2  and  3  o'clock  are  the  hands  of 
a  clock  30  minute  spaces  apart  ? 

5.  What  angle  is  formed  by  the  hands  of  a  clock  at  2.30? 
*6.  At  what  time  between  5  ,and  6  o'clock  are  the  hands  of 

a  clock  20  min.  apart  ?    How  many  answers  ?    How  may  these 
results  be  obtained  from  the  formula  of  Ex.  2  ? 

326.  Work  problems.  The  work  problem  is  another 
type  of  problem  easily  solved  by  formula. 

EXERCISES 

1.  One  pipe  will  fill  a  tank  in  3  hr.  and  a  second  pipe  can 
fill  it  in  4  hr.    How  long  will  it  take  to  fill  the  tank  if  both 
pipes  are  left  running  ? 

Let      n  =  the  number  of  hours  it  will  take  both  pipes  to  fill  the  tank. 

Then  —  =  the  part  of  the  tank  filled  in  1  hr., 
n 

i  =  the  part  of  the  tank  filled  by  the  first  pipe  in  1  hr., 
and  ^  =  the  part  of  the  tank  filled  by  the  second  pipe. 

Hence  -  +'-  =  -•  Why? 

3       4       n 

Multiplying  by  12  n,        4  n  +  3  n  =  12, 
or  7  n  =  12. 

Whence  n  =  If  hr. 

2.  A  can  lay  a  drain  in  5  da.  and  B  can  do  it  in  7  da.   How 
long  will  it  take  both  working  together  ? 

3.  One  boy  can  drive  his  car  over  a  trip  in  8  hr.  and  a 
second  boy  can  make  the  trip  in  5  hr.    How  long  would  it  take 
them  to  meet  if  each  started  at  an  end  ? 

NOTE.  It  is  clear  from  the  foregoing  problems  that  any  numbers 
would  be  used  just  as  3  and  4  are  used  in  Ex.  1.  Hence  a  formula 
may  be  obtained,  as  is  shown  by  Ex.  4. 


286  GENERAL  MATHEMATICS 

4.  A  can  do  a  piece  of  work  in  a  days  and  B  can  do  it  in 
b  days.    How  long  will  it  take  them  to  do  it  together  ? 

Let  n  =  the  number  of  days  it  will  take 

them  together. 

Then  -  =  the  amount  of  work  they  can  do 

n  ,    , 

in  1  da., 

-  =  the  amount  A  can  do  in  1  da., 
a 

and  -  =  the  amount  B  can  do  in  1  da. 

b 

1,1      1 
Hence  _  +  _  —  _. 

a      b      n 

Multiplying  by  aim,        l>n  +  an  =  ah, 
(J>  +  a)n  —  ah, 
ab 


a  +  I > 
NOTE.   Any  problem  of  the  type  of  Ex.  1  on  page  285  may  be 

solved  by  using  the  equation  n  = as  a  formula.    Thus,  to  solve 

a  +  b 

Ex.  1  let  a  =  3,  b  —  4.    Then  n  =  — —  =  —  =  1^  hr. 

5.  One  boy  can  make  a  paper  route  in  2  hr.  and  his  friend 
can  make  the  route  in  1^  hr.  How  long  will  it  take  the  two 
together  ?  (Solve  by  formula.) 

*6.  Suppose  that  in  Ex.  1  on  page  285  the  second  pipe  is  an 
emptying  pipe,  how  long  will  it  take  to  fill  the  tank  if  both 
pipes  are  running  ?  What  form  does  the  formula  take  ? 

*7.  A  can  sweep  a  walk  in  7  min.,  B  in  8  min.,  and  C  in 
10  min.  How  long  will  it  take  them  working  together  ?  What 
form  does  the  formula  for  a  work  problem  take  ? 

*8.  A  could  lay  a  sidewalk  in  3  da.,  B  in  4  da.,  and  C  in 
4.5  da.  How  long  does  it  take  them  when  working  together  ? 
Solve  by  substituting  in  the  formula  for  Ex.  7. 


CONTROL  OF  THE  FORMULA  287 

327.  Translating  rules  of  procedure  into  formulas.  Write 
each  of  the  following  in  the  form  of  a  formula : 

1.  The  area  of  a  triangle  equals  the  product  of  half  the 
base  times  the  altitude. 

2.  The  area  of  a  rectangle  equals  the  product  of  its  base 
and  altitude. 

3.  The  area  of  a  parallelogram  equals  the  product  of  its 
base  and  altitude. 

4.  The  area  of  a  trapezoid  equals  one  half  the  sum  of  the 
parallel  bases  multiplied  by  the  altitude. 

5.  The  volume  of  a  pyramid  equals  one  third  the  base 
times  the  altitude. 

6.  The  length  of  a  circle  is  approximately  equal  to  twenty- 
two  sevenths  of  the  diameter. 

7.  The  circumference  of  a  circle  is  equal  to  TT  times  the 
diameter. 

8.  The  area  of  a  circle  is  TT  times  the  square  of  the  radius. 

9.  The  product  equals  the  multiplicand  times  the  multiplier. 

10.  The  product  obtained  by  multiplying  a  fraction  by  a 
whole  number  is  the  product  of  the  whole  number  and  the 
numerator  divided  by  the  denominator. 

11.  The  quotient  of  two  fractions  equals  the  dividend  multi- 
plied by  the  inverted  divisor. 

12.  The  square  root  of  a  fraction  equals  the  square  root  of 
the  numerator  divided  by  the  square  root  of  the  denominator. 

13.  The  square  of  a  fraction  is  the  square  of  the  numerator 
divided  by  the  square  of  the  denominator. 

14.  The  rule  for  calculating  the  cost  of  one  article  when 
you  know  that  a  certain  number  of  them  cost  so  much ;  write 
the  cost  of  m  articles. 

15.  The  rule  for  expressing  years,  mouths,  and  days  as  years. 


288  GENERAL  MATHEMATICS 

16.  The  rule  for  calculating  the  area  of  three  adjacent  rooms 
of  different  lengths  but  the  same  width. 

17.  The  rule  for  calculating   the   area  of   the   floor  of  a 
square  room. 

18.  The  rule  for  finding  the  cost  of  a  telegram. 

19 .  The  rule  for  finding  the  area  of  a  figure  cut  from  cardboard, 
given  its  weight  and  the  weight  of  a  square  unit  of  cardboard. 

20.  The  rule  for  finding  the  amount  of  available  air  for  each 
person  in  a  classroom,  given  the  dimensions  of  the  room  and 
the  number  in  the  class. 

21.  The  rule  for  finding  the  weight  of  a  single  lead  shot, 
given  the  weight  of  a  beaker  with  a  given  number  of  shot  in 
it  and  the  weight  of  the  empty  beaker. 

22.  The  rule  for  predicting  the  population  of  a  town  after 
a  certain  number  of  months,  given  the  present  population,  the 
average  number  of  births,  and  the  average  number  of  deaths. 

23.  The  rule  for  finding  the  distance  apart,  after  a  given 
time,  of  two  cars  which  start  from  the  same  point  and  travel 
in  opposite  directions  at  different  speeds. 

24.  The  same  as  in  Ex.  23  except  that  the  cars  are  m  miles 
apart  at  starting. 

25.  The  same  as  in  Ex.  23  except  that  the  cars  go  in  the 
same  direction  with  different  speeds. 

26.  The  reading  on  a  Fahrenheit  thermometer  is  always  32° 
greater  than  ^  of  the  reading  on  a  centigrade  thermometer. 

27.  The  reading  of  a  centigrade  thermometer  may  be  cal- 
culated by  noting  the  reading  on  the  Fahrenheit,  subtracting 
32,  and  taking  |  of  this  result. 

328.  Graphic  representation  of  the  relation  between  the 
readings  on  centigrade  and  Fahrenheit  thermometers.  The 
last  two  exercises  deal  with  two  types  of  thermometers 
that  are  used  to  measure  temperature.  Fig.  207  shows  that 


CONTROL  OF  THE  FORMULA 


289 


F 


Boiling  100' 
90f 
80' 
70' 
60' 
50' 
40' 
30' 
20' 
10' 

Freezing  0' 

-10' 

-17.78' 


the  only  fundamental  difference  is  the  different  graduations 
of  the  scale.  In  the  Fahrenheit  thermometer  the  place 
where  the  mercury  stands  if  immersed  in  freezing  water  is 
32°  ;  on  the  centigrade  it  is  zero.  This  defines  the  freezing 
point  on  each.  The  boiling  point  is 
marked  212°  on  the  Fahrenheit  and 
only  100°  011  the  centigrade.  Hence 
the  Fahrenheit  thermometer  has  180 
divisions  of  the  scale  in  the  interval 
from  freezing  to  boiling,  while  the 
centigrade  has  but  100.  This  means 
that  a  unit  on  the  centigrade  is 
longer,  or  for  any  space  on  the  centi- 
grade there  are  i|$  times,  or  |  times, 
as  many  Fahrenheit  units.  Hence 
the  number  of  units  in  a  Fahrenheit 
reading  equals  -|  times  as  many  cen- 
tigrade units  plus  32°  (which  are 
below  the  freezing  point).  Stated 
as  a  formula,  F=fC  +  32. 

We  could,  of  course,  always  trans-     BETWEEN      THE      CENTI- 

late  the  reading  on  one  thermometer    GRAI)E  AND  FAHRENHEIT 

...  ,.  THERMOMETERS 

to  the  corresponding  reading  on  the 

other  by  means  of  this  formula.  It  is  far  easier,  how- 
ever, to  graph  the  linear  equation  F  =  |C  +  32,  which 
will  reveal  possible  relations,  as  is  shown  in  Fig.  208. 

NOTE.    The  pupil  should  construct  the  graph  independently  of 
the  text,  using  a  table  similar  to  the  following : 

C.        F. 

Let  C.=    0,  then  F.  =  32. 

Let  C.  =  25,  then  F.  =  77. 

Let  C.  =  15,  then  F.  =  59. 


212° 

194° 

-176° 

-158° 

140° 

-122° 

-104° 

86° 

68° 

50° 

32° 

14° 

0° 


FIG.  207.   THE  RELATION 


290 


CKNKRAL   MATHEMATICS 


EXERCISES 

1.  Determine  by  the  graph  the  corresponding  Fahrenheit 
readings  for  the  following  centigrade  readings :    5°,  10°,  20°, 
30°,  -  5°,  -  10°,  -  15°,  -  25°. 

2.  Determine  by  the  graph  the  centigrade    readings   cor- 
responding to  the  following  Fahrenheit   readings :    80°,  70°, 
60°,  30°,  20°,  10°,  -  5°,  -  10°. 

3.  In  the  formula  F  =  f  C 
-f-  32°  substitute  in  each  case 
the  two  numbers  you  think  are 
corresponding  readings.    The 
error  should  be  very  small. 

4.  Normal    room    tempera- 
ture   is    68°  F.     What    is    it 
centigrade  ? 

5.  The  normal  temperature 
of  the  human  body  is  98.4°  F. 
What  is  it  centigrade  ? 

6.  What  temperature  centi- 
grade corresponds  to  0°  F.? 


7.  Could  you  go  skating  at     FlG  20g    A  GRAPH  T0  BE  rSK1)  JN 

15°  C.  ?  CHANGING    CENTIGRADE    READINGS 

TO    FAHRENHEIT  AND   VICE   VERSA 

8.  In  your  general-science 

course  you  are  told  that  mercury  freezes  at   —  40°  F.  What 
is  this  centigrade  ? 

9.  Would  your  classroom  be  comfortable  at  25°  C.  ? 

329.  Evaluating  a  formula.  Find  the  value  of  the  letter 
called  for  in  each  of  the  exercises  given  on  page  291. 
When  no  explanation  is  given,  it  is  assumed  that  the 
student  recognizes  the  formula. 


CONTROL  OF  THE  FORMULA 


291 


EXERCISES 

1.  Given    C  =  |    (F  -  32).      Find    C.    if    F.  =  0°;    32°; 
212°;   100°. 

2.  Given     F  =  |C.+  32.      Find    F.    if    C.  =  0°;     100°; 
-20°;  60°. 

3.  Given  d  =  rt.    Find   d   if   /•  =  87.5  mi.  per  hour  and 
t  =  12  hr.  ;  if  r  =  10^  ft.  per  second  and  t  —  10  sec. 

4.  Given  r  =  -  •    Find  r  if  d  =  1  mi.  and  t  =  4  min.  16  sec.  ; 

{/ 

if  d  =  ^  mi.  and  t  =  2.07  sec. 

5.  Given  A=P+  Prt.    Find  A  if  P  =  $240,  r  =  4-}%,  and 
*=lyr.  2  mo.  3  da.;  if  P  =  $128,  r=6%,  and  £  =  2yr.  3  da.; 
if  P  =  $511,  r  =  6%,  and  t  =  20  yr. 

6.  Given  V=lwh.  Find  F  (see  Fig.  209) 
if  I  =12.2  ft.,  w  =  8.3  ft.,  and  h  =  6.4  ft.; 
if  £  =  9.3  in.,  iv  =  5.6  in.,  and  h  =  1  in. 


7. 


Find  Tiff  =63  ft., 


«  FIG.  209.    RECTAX- 

w  =  2.4  ft.,   and  h  =  1.6  ft.;    if   I  —  3  cm.,       GULAR  PARALLELE- 
iv  =  2.1  cm.,  and  h  =  1.4  cm.  PIPED 

8.  Given  c  =  ^-  d.    Find  c  if  d  =  1  f t. ;  1  in. ;  4  in. ;  10  in. ; 
5-1  in. 

9.  Given  A  =  Trr2  (TT  =  -2T2-).     Find  .4   if  r  =  lin.;    5ft.; 
10  yd. ;  7  m. ;  8.5  cm. 

10.  The  volume  F  of  any  prism  (Fig.  210) 
is  equal  to  the  product  of  its  base  B  and 
its  altitude  h.   Find  V  if  B  =  246.12  sq.  in. 
and  h  =  12 in.;  if  B  =  212.44  sq.  in.  and 
&  =  2|  ft. 

11.  The    lateral    surface  L  of  a  right 

prism   equals   the  perimeter  of  the  base        FlG  2io.  PRISM 

P  times  the  altitude  h.    Find    L   if  P  = 

126  in.    and    h  =  11  in.;    if   P  =  21.6  in.    and   h  =  0.35  in. 


292 


GENERAL  MATHEMATICS 


CYLINDER 


12.  The  volume  of  a  right  cylinder  (Fig.  211)  is  equal  to 
the  product  of  its  base  and  height.    The  formula  is  V  =  TrrVi, 
where  r  is  the  radius  of  the  circular  base.  Find  Vii  r=10.2  cm. 
and  h  =  9  cm. ;   if  r  =  6  in.  and   h  —  12  in. 

13.  The  lateral  surface  of  a  right  cylinder 
equals  the  product  of  the  altitude  and  the 
circumference    of   the    base.     The    formula 
usually  given  is  S  =  Ch.   Find  S  if  C  =  *f  ^  in. 
and  h  =  10  in. 

14.  The  entire  surface  T  of  a  right  cylinder 
equals  the  circumference  of  the  circular  base 

times  the  sum  of  the  altitude  and  the  radius  of  Fj(.     2n 
the  base  ;  that  is,  T  =  2  irr(r  +  h).   Find  T  if 
A  =10  in.  and  r  =  5  in.;  if  A  =  2ft.  and  r=l  ft. 

15.  The  volume  V  of  any  pyramid  (Fig.  212) 
is  equal  to  one  third  the  product  of  its  base 

B  and  its  altitude  h ;  that  is,  V  =  —  •    Find 

o 

V  if  £  =  200sq.  in>  and  A  =12  in.;  if  B  = 
24.6  sq.  in.  and  h  =  2  ft. 

16.  The  lateral  area  S  of  a  regular  pyramid 

is  equal  to  one  half  the  product  of  the  perimeter  P  of  its  base 

Pi 

and  its  slant  height  £;  that  is,  S  =  —  •    Find  A  if  P  =  10. 6  in. 

£t 

and  I  =  8.2  in. ;  if  P  =  4.3  cm.  and  I  =  15  cm. 

17.  The  lateral  area  S  of  a  right  circular 
cone  (Fig.  213)  is  equal  to  one  half  the  prod- 
uct of  its  slant  height  I  and  the  circumfer- 
ence C  of  its  base.    Write  the  formula  for  S, 
and  find  S  if  1=  14.6  in.  and  C  =  10in. ;  if 

I  =  3.6  ft.  and  C  =  31.416  ft.  FIG.  213.   RIGHT 

18.  The  lateral  area  S  of  a  right  circular  cone     ClRCULAR  CONE 
is  Trrh,  where  r  is  the  radius  of  the  base  and  h  is  the  slant  height. 
Find  S  if  r  =  10  in.  and  h  =  10  in. ;  if  r  =  10  in.  and  h  =  26.2  in. 


FIG.  212.  PYRAMID 


CONTROL  OF  THE  FORMULA  293 

19.  The  entire  surface  T  of  a  right  circular  cone  equals  the 
lateral  area  I  plus  the  area  of  the  base ;  that  is,  T  =  trrl  +  Trr2, 
or  7rr(l  +  r).   Find  T  if  I  =  10  in.  and  r  =  5  in. ;  if  I  =  12.6  in. 
and  r  =  6  in. 

20.  An  object  falling  from  rest  falls  in  a  given  time  a 
distance  equal  to  the  product  of  16  and  the  square  of  the 
number  of  seconds  it  has   fallen;    that  is,  d  =  16t2.     Find 
d  if  t  =  1  sec. ;  2  sec. ;  3  sec. ;  4  sec. 

21.  An  object  thrown  downward  travels  in  a  given  time  a  dis- 
tance equal  to  the  product  of  16  and  the  square  of  the  number 
of  seconds  it  has  fallen,  plus  the  product  of  the  velocity  with 
which  it  is  thrown  and  the  number  of  seconds  it  is  falling.   The 
formula  is  S  =  16 1*  +  Vt.  Find  S  if  t  =  3  sec.  and  V=  13  ft.  per 
second;  if  t  =  5 sec.  and  F=100ft.  per  second. 

22.  The  volume  of  a  sphere  (Fig.  214)  equals 
the  cube  of  the  radius  multiplied  by  ^  TT.    Find  V 
if  y  =  1  in.;  10  in.;  5  in.;  10ft.;  12ft. 

23.  The  surface   S  of  a  sphere  is  equal  to    FIG.  214.  THF, 
4  7H-2.    Find  S  if  r  =  10  in. ;  12  ft. ;  6j  ft. 

*24.  The  force  of  pressure  P  of  the  wind,  in  pounds  per 
square  foot,  is  given  by  the  equation  P  =  0.005  F2,  where  V  is 
the  velocity  pf  the  wind  in  miles  per  hour.  What  would  be  the 
total  pressure  of  this  wind  against  the  side  of  a  wall  25  ft. 
high  and  80  ft.  long  of  a  wind  blowing  30  mi.  per  hour  ? 

*25.  Show  that  the  formula  for  the  length  I  of  a  belt  passing 
around  two  pulleys  of  the  same  size  whose  radii  are  each  r  feet, 
and  the  distance  between  whose  centers  is  d  feet,  is  1=  2  7rr+  2  d. 
Find  I  when  r  =  l£  and  d  =  4|. 

26.  In  a  price  list  the  cost  of  sewer  pipe  per  foot  of  length 
is  given  by  the  formula  C  =  0.4  d2  + 14,  where  d  is  the  diam- 
eter of  the  pipe  in  inches  and  C  the  cost  in  cents.  What  will 
be  the  cost  of  20  ft.  of  pipe  2  in.  in  diameter  ? 


294  GENERAL  MATHEMATICS 

330.  Practice  in  solving  for  any  letter.  It  is  often  desir- 
able to  solve  a  formula  for  some  particular  letter  in  that 
formula.  Too  often  the  student  will  recognize  a  formula 
provided  it  stands  in  the  form  in  which  it  is  commonly 
written,  but  will  not  appreciate  its  meaning  if  it  is  written 
in  a  different  way.  For  example,  how  many  students 

V 

would   recognize  the  formula   c  =  —  as   the   well-known 

formula  V=abc?  It  is  the  same  formula  except  that  it 
is  in  a  different  form.  If  the  student  realizes  this,  it  helps 
him  to  gain  control  of  the  formula.  The  following  exer- 
cises will  furnish  practice  in  solving  for  particular  letters. 

EXERCISES 

Solve  each  of  the  following  formulas  for  the  letter  or  letters 
indicated : 

ab  Bh 

1.  .-I  =  —  for  a  ;  for  />.  11.  T  =  —  for  h. 

2.  r  =  aJcforc;  fora;  forb.  Pi. 

12.  .1  =  —  for  /. 

3.  <?  =  rtforr;  for  t. 

D  13.  A  =  P  +  Prt  for  t. 

4.  ^  =  Yfor,/;for/>.  14.  r  =  |(F- 32)  for  F. 

5.  WA*WJJ*to**Wt        15     Sss9*fmi,.  lory. 


6-  V-    —  for  w;  for  A.  i6.   s  =  2  irrh  +  2  ir>*  for  h. 

7.  C  =  2.5  (7  for  r/.  17.   .1  =     l    9  '2 —  for  //  ; 

8.  r  =  41 ,.  for  r.  for  ^  ;  for  6a. 

9.  V  =  Bh  for  B.  18.  c  =  _A_  for  E  .  for  ;,  . 
10.  V=*£i*h  for  A.  for  r. 


CONTROL  OF  THE  FORMULA  295 

SUMMARY 

331.  This  chapter  has  taught  the  meaning  of  the  following 
words  and  phrases :  formula,  solving  a  formula,  evaluating 
a  formula,  applying  a  formula,  centigrade,  Fahrenheit. 

332.  A  formula  is  a  conveniently  abbreviated  -form  of 
some  practical  rule  of  procedure. 

333.  A  clear  understanding  of  a  formula  implies: 

1.  An  analysis  of  some  arithmetical  situation  so  as  to 
arrive  at  some  rule  of  procedure. 

2.  Translating  the  rule  into  a  formula. 

3.  The  ability  to  solve  for  any  letter  in  terms  of  the 
other  letters  in  the  formula. 

4.  The    ability  to    apply  the    formula   to  a  particular 
problem  and  to  evaluate  the  formula. 

334.  The  preceding  steps  were  illustrated  in  detail  by 
applications  to  interest  problems,  to  problems  involving 
motion,  to  work  problems,  to  thermometer  problems,  and 
to  geometric  problems. 

335.  The  graphical  interpretations  suggested  economical 
methods  of  manipulating  a  formula.    For  example : 

1.  Simple-interest  problems  were  solved  by  the  formulas 
1  =  Prt  and  A  =  P  +  Prt. 

2.  A  problem  involving  uniform  motion  in  a  straight 
line  was  solved  by  the  formula  d  =  rt. 

3.  The    relation    between    centigrade    and    Fahrenheit 
readings  was  expressed  by  the  formula  (7  =!(/*— 32). 

336.  While  the  important  thing  in  this  chapter  is  the 
power  of  manipulating  and  evaluating  a  formula,  the  stu- 
dent was  given  the  meaning  of  most  of  the  formulas  in 
order  to  have  him  realize  from  the  very  outset  that  both  the 
formulas  and  their  manipulation  refer  to  actual  situations. 


296  GENERAL  MATHEMATICS 

HISTORICAL  NOTE.  The  development  of  the  formula  belongs  to 
a  very  late  stage  in  the  development  of  mathematics.  It  requires  a 
much  higher  form  of  thinking  to  see  that  the  area  of  any  triangle 

can  be  expressed  by  A  =  —  than  to  find  the  area  of  a  particular 

A 

lot  whose  base  is  two  hundred  feet  and  whose  altitude  is  fifty  feet. 
Hence,  it  was  very  late  in  the  race's  development  that  letters  were 
used  in  expressing  rules. 

The  early  mathematicians  represented  the  unknown  by  some 'word 
like  res  (meaning  "  the  thing  ").  Later,  calculators  used  a  single  letter 
for  the  unknown,  but  the  problems  still  dealt  with  particular  cases. 
Diophantus,  representing  Greek  mathematics,  stated  some  problems 
in  general  terms,  but  usually  solved  the  problems  by  taking  special 
cases.  Vieta  used  capital  letters  (consonants  and  vowels)  to  represent 
known  and  unknown  numbers  respectively.  Newton  is  said  to  be 
the  first  to  let  a  letter  stand  for  negative  as  well  as  positive  numbers, 
which  greatly  decreases  the  number  of  formulas  necessary. 

While  the  race  has  had  a  difficult  time  discovering  and  under- 
standing formulas,  it  takes  comparatively  little  intelligence  to  use  a 
formula.  Many  men  in  the  industrial  world  do  their  work  efficiently 
by  the  means  of  a  formula  whose  derivation  and  meaning  they  do 
not  understand.  It  is  said  that  even  among  college-trained  engineers 
only  a  few  out  of  every  hundred  do  more  than  follow  formulas  or 
other  directions  blindly.  Thus,  it  appears  that  for  the  great  majority 
only  the  immediately  practical  is  valuable.  However,  we  can  be 
reasonably  sure  that  no  one  can  rise  to  be  a  leader  in  any  field  by 
his  own  ability  without  understanding  the  theoretical  as  well  as 
the  practical. 

The  formula  is  very  important  in  the  present  complex  industrial 
age.  A  considerable  portion  of  the  necessary  calculation  is  done  by 
following  the  directions  of  some  formula.  Therefore  to  meet  this 
need  the  study  of  the  formula  should  be  emphasized.  In  discussing 
the  kind  of  mathematics  that  should  be  required  Professor  A.  R. 
Crathorne  (School  and  Society,  July  7,  1917,  p.  14)  says:  "Great 
emphasis  would  be  placed  on  the  formula,  and  all  sorts  of  formulas 
could  be  brought  in.  The  popular  science  magazines,  the  trade 
journals  and  catalogues,  are  mines  of  information  about  which  the 
modern  boy  or  girl  understands.  The  pupil  should  think  of  the 
formula  as  an  algebraic  declarative  sentence  that  can  be  translated 


ARCHIMEDES  (287-212  B.C.) 
(Bust  in  Naples  Museum) 


298  GENERAL  MATHEMATICS 

into  English.  The  evaluation  leads  up  to  the  tabular  presentation 
of  the  formula.  Mechanical  ability  in  the  manipulation  of  symbols 
should  be  encouraged  through  inversion  of  the  formula,  or  what  the 
Englishman  calls  'changing  the  subject  of  the  formula.'  We  have 
here  also  the  beginning  of  the  equation  when  our  declarative  sentence 
is  changed  to  the  interrogative." 

Archimedes  (287-212  H.C.),  a  great  mathematician  who  studied 
in  the  university  at  Alexandria  and  lived  iu  Sicily,  loved  science 
so  much  that  he  held  it  undesirable  to  apply  his  information  to 
practical  use.  But  so  great  was  his  mechanical  ability  that  when 
a  difficulty  had  to  be  overcome  the  government  often  called  on 
him.  He  introduced  many  inventions  into  the  everyday  lives  of 
the  people. 

His  life  is  exceedingly  interesting1.  Read  the  stones  of  his  detec- 
tion of  the  dishonest  goldsmith  ;  of  the  use  of  burning-glasses  to 
destroy  the  ships  of  the  attacking  Roman  squadron ;  of  his  clever 
use  of  a  lever  device  for  helping  out  Hiero,  who  had  built  a  ship 
so  large  that  he  could  not  launch  it  off  the  slips ;  of  his  screw  for 
pumping  water  out  of  ships  and  for  irrigating  the  Nile  valley.  He 
devised  the  catapults  which  held  the  Roman  attack  for  three  years. 
These  were  so  constructed  that  the  range  was  either  long  or  short 
and  so  that  they  could  be  discharged  through  a  small  loophole 
without  exposing  the  men  to  the  fire  of  the  enemy. 

When  the  Romans  finally  captured  the  city  Archimedes  was  killed, 
though  contrary  to  the  orders  of  Marcellus,  the  general  in  charge  of 
the  siege.  It  is  said  that  soldiers  entered  Archimedes'  study  while  he 
was  studying  a  geometrical  figure  which  he  had  drawn  in  sand  on 
the  floor.  Archimedes  told  a  soldier  to  get  off  the  diagram  and  not 
to  spoil  it.  The  soldier,  being  insulted  at  having  orders  given  to 
him  and  not  knowing  the  old  man,  killed  him. 

The  Romans  erected  a  splendid  tomb  with  the  figure  of  a  sphere 
engraved  on  it.  Archimedes  had  requested  this  to  commemorate  his 
discovery  of  the  two  formulas :  the  volume  of  a  sphere  equals  two- 
thirds  that  of  the  circumscribing  right  cylinder,  and  the  surface  of 
a  sphere  equals  four  times  the  area  of  a  great  circle.  You  may  also 
read  an  interesting  account  by  Cicero  of  his  successful  efforts  to 
find  Archimedes'  tomb.  It  will  be  profitable  if  the  student  will 
read  Ball's  "  A  Short  History  of  Mathematics,"  pp.  65-77. 


CHAPTER  XII 

FUNCTION ;  LINEAR  FUNCTIONS  ;  THE  RELATED  IDEAS  OF 

FUNCTION,     EQUATION,     AND     FORMULA     INTERPRETED 

GRAPHICALLY;   VARIATION 

337.  Function  the  dependence  of  one  quantity  upon 
another.  One  of  the  most  common  notions  in  our  lives  is 
the  notion  of  the  dependence  of  one  thing  upon  another. 
We  shall  here  study  the  mathematics  of  such  dependence 
by  considering  several  concrete  examples. 

EXERCISES 

1.  Upon  what  does  the  cost  of  10yd.  of  cloth  depend? 

2.  If  Resta  drives  his  car  at  an  average  rate  of  98.3  mi.  per 
hour,  upon  what  does  the  length  (distance)  of  the  race  depend  ? 

3.  A  boy  rides  a  motor  cycle  for  two  hours.    Upon  what 
does  the  length  of  his  trip  depend  ? 

4.  How  much  interest  would  you  expect  to  collect  in  a 
year  on  $200? 

5.  Upon  what  does  the  length  of  a  circular  running  track 
depend  ? 

6.  A  man  wishes  to  buy  wire  fencing  to  inclose  a  square  lot. 
How  much  fencing  must  he  buy  ? 

7.  State  upon  what  quantities  each  of  the  following  depends : 

(a)  The  amount  of  sirloin  steak  that  can  be  bought  for  a  dollar. 

(b)  The  number  of  theater  tickets  that  can  be  bought  for 
a  dollar. 

(c)  The  height  of  a  maple  tree  that  averages  a  growth  of 
4  ft.  per  year. 

299 


300  GENERAL  MATHEMATICS 

(d)  The  time  it  takes  you  to  get  your  mathematics  lesson  if 
you  solve  one  problem  every  three  minutes. 

(e)  The  value  of  a  submarine  as  a  merchant  vessel. 

(f)  The  rate  of  interest  charged  by  your  local  bank. 

(g)  The  perimeter  4  x  —  4  of  the  rectangle  in  Fig.  215. 

The  preceding  exercises  illustrate  the  dependence  of  one 
quantity  upon  another.  We  have  had  numerous  other 
examples  of  dependence  in  the  chapters  on  statistics  and 
the  formula.  In  fact,  every  practical  formula  implies  that 
the  value  of  some  quantity  depends  upon  one  or  more 
others.  Thus  the  circumference  of  a  circular  running  track 
depends  upon  the  diameter.  When  a  quantity  depends 
upon  another  quantity  for  its  value,  it  is  said  to  be  a  func- 
tion of  the  latter.  Thus  the  area  of  a  circle  is  a  function 
of  the  diameter  because  it  depends  upon  the  diameter  for 
its  value ;  the  amount  of  sirloin  steak  that  can  be  bought 
for  a  dollar  is  a  function  of  the  price  per  pound ;  and  the 
expression  4  x  —  4  is  a  function  of  x  because  its  value 
changes  with  every  change  in  the  value  of  x. 

See  if  you  can  illustrate  the  idea  of  function  by  ten 
familiar  examples  not  given  above. 

338.  Variable.    A  number  that  may  change,  assuming  a 
series  of  values  throughout  a  discussion,  is  called  a  variable. 
It  is  not  obliged  to  vary — it  is  "  able  to  vary."  Thus  the  price 
of  wheat  and  the  number  s  in  the  equation  A  =  s2  are  variables. 

339.  Dependent  and  independent  variables.  In  the  formula 
C  =  ird  the  number  d  is  said  to  be  the  independent  variable. 
In  a  discussion  or  in  the  construction  of  circles  we  may 
take  its  value  equal  to  any  number  we  please.     On  the 
other  hand,  the  value  of  C  is  automatically  fixed  once  the 
value  of  d  is  determined.    Because  of  this  fact  C  is  called 
the  dependent  variable. 


FUNCTION  INTERPRETED  GRAPHICALLY    301 

EXERCISES 

1.  What  is  the  value  of  C  in  the  equation  C  —  ird  if  d  —  2? 
if  d  =  5?  if  d  =  W? 

2.  Illustrate  the  ideas  of  dependent  and  independent  vari- 
ables with  examples  chosen  from  the  text  or  from  your  own 
experience. 

340.  Constant.    The  number  TT  in  the  formula  C  =  Trd 
differs  from  C  and  d  inasmuch  as  it  never  changes  at  any 
time  in  the  discussion  of  circles.    This  number  is  approxi- 
mately -2y2-,  or  3.1416,  whether  we  are  dealing  with  small  or 
large  circles.    We  therefore  call  a  number  like  this,  which 
has  a  fixed  value,  a  constant.    Obviously  any  arithmetical 
number  appearing  in  a  formula  is  a  constant ;  thus  the  2  in 

A  =  —  and  the  •§-  and  the  32  in  .7^=  ^  £7+  32  are  constants. 
2 

EXERCISE 

Turn  to  Chapter  XI,  on  the  formula,  and  find  five  formulas 
that  illustrate  the  idea  of  a  constant. 

341.  Graph  of  a  function.    A  graph  may  be  constructed 
showing  how  a  function    changes   as   the    value    of    the 
independent    variable    changes.  x 

The  rectangle  in  Fig.  215  is  a 

picture  (either  enlarged   or  re-    x-z  x-z 

duced)  of  every  rectangle  whose 
length  exceeds  its  width  by  two 
units.  We  shall  now  proceed  to  FlG  2i5 

show  graphically  that  the  perim- 
eter varies  with  every  change  in  the  value  of  x.    The  table 
on  the  following  page  gives  the  corresponding  values  for 
the  length  x  and  for  4  x  —  4,  the  perimeter. 


302 


GENERAL  MATHEMATICS 


X 

3 

4 

5 

6 

7 

8 

9 

10 

4x-4 

8 

12 

16 

20 

24 

4x-4 


--20 


If  we  plot  the  points  corresponding  to  (3,  8),  (4,  12), 
(5, 16),  etc.,  using  the  horizontal  axis  to  plot  the  values  of  x 
and  the  vertical  axis  to  plot  the  values  of  4  x  —  4,  we  obtain 
the  points  as  shown  on  the  straight  line  AB  in  Fig.  216.  The 
line  shows  that  as  x  increases,  the  value  of  4  x  —  4  increases 

accordingly. 

EXERCISES 

1.  Tell  in   your  own   words   how   the  graph  in  Fig.  216 
shows    that   the   function   4  x  —  4   increases    as  x  increases. 

2.  Determine  from  the  graph 
the     perimeters      of  .  rectangles 
whose   lengths    are   as    follows : 
8  in.;, 9  in.;  10.5  in. ;   11  in. 

3.  Determine  from  the  graph 
the    length    of    the    rectangles 
whose  perimeters  are  as  follows  : 
30  in.;    25  in.;    20  in. ;     18  in.; 
10  in. ;  3  in. ;  0  in. 

4.  Suppose  you  chose  to  make 
a  particular  rectangle  10  in.  long. 
How  long  would  the  perimeter  be? 
How  does  the  graph  show  this  ? 

5.  How  long  would  you  make 
a  rectangle  of  the  same  shape 
as  the  one  in  Fig.  215  so  as  to 
have  its  perimeter  16  in.  ?    How 
does  the  graph  show  this  ? 

6.  Relying  on  your  past  experience,  tell  how  many  rectangles 
you  could  construct  in  the  shop  or  construct  in  your  notebook 
"whose  length  shall  exceed  their  width  by  two  units  ". 


FIG.  216.    GRAPH  SHOWING  THAT 
THE  PERIMETER  OF  THE  RECTAN- 
GLE IN  FIG.  215  is  A  FUNCTION 
OF  THE  LENGTH 


FUNCTION  INTEEPBETED  (GRAPHICALLY    303 

342.  Linear  function.    Since  the  graph  of  the  expression 
4  x  —  4  is  a  straight  line,  the  function  is  called  a  linear  func- 
tion.   If  we  let  y  represent  the  value  of  the  linear  function, 
we  get  the  corresponding  linear  equation  y  —  4  jc  —  4. 

EXERCISE 

Give  five  examples  of  linear  functions. 

343.  Solving  a  family  of   equations   by  means  of   the 
graph.    The'  graph  of  the  function  4  x  —  4  may  be  used 
to  solve  all  equations  one  of  whose  members  is  4  r  —  4 
and  the  other  some  arithmetical  number  or  constant.    For 
example,   if   in    the    equation  y  =  4x  —  4  we    let  ;z/  =  16, 
then  the  equation  4  #  —  4  =  16  may  be  interpreted  as  rais- 
ing the  question,  What  is  the  value  'of  x  that  will  make 
4. r  —  4=16?    In  order  to  answer  this  question  we  find 
16  on  the  ?/-axis  (the  vertical  axis),  pass  horizontally  to 
the  graph  of  4  x  —  4,  and  read  the  corresponding  value 
of  x.   The  corresponding  value  of  x  is  seen  to  be  5.   Hence 
4  x  —  4  =  16  when  x  =  5. 

As  a  verbal  problem  the  equation  4 a*— 4=16  may  be 
translated  into  the  following  interrogative  sentence:  What 
shall  be  the  length  of  the  rectangle  in  order  that  it  may 
have  a  perimeter  of  16  ?  A  glance  at  the  graph  is  sufficient 
to  determine  the  answer ;  namely,  5. 

EXERCISES 
Solve  by  graph,  and  check  the  following  equations  : 

1.  4z-4  =  20.  5.  4z-8  =  2. 

0  HINT.    Add  4  to  both  mem- 

2.  4ic  —  4  =  Z4.  . 

bers  so  as  to  obtain  the  equation 

3.  4.r  -  4  =  12.  4z-4  =  6. 

4.  4»-  4  =  6.  6.  4x-5  =  13. 


304  <;KM-:KAL  MATHEMATICS 

7.  4.r-9  =  10.  9.  4ir+  2  =  12. 

8.  4.r  +  6  =  26.  10.  4a:  +  5  =  19. 

HIM.    Subtract  10  from  both  -  -  J.    •  _i_  1  <"  _  <><i 
members    so    as   to   obtain    the 

result  4  x  -  4  =  16.  12.  4  ./•  -  4  =  0. 

344.  The  graphical  solution  of  the  function  set  equal  to 
zero.  Problem  12,  Art.  343,  is  an  interesting  special  case 
for  two  reasons:  (1)  It  gives  us  an  easy  method  of  find- 
ing the  value  of  x  in  the  equation  4^  —  4  =  0.  We  need 
only  refer  to  the  graph  and  observe  where  the  line  crosses 
the  o>axis.  The  line  is  seen  to  cross  where  #  =  1.  This  value 
of  x  checks  because  4-1—4  =  0.  Hence  x  =  1  is  a  solu- 
tion of  the  equation  4  x  —  4  =  0.  (2)  It  furnishes  us  a 
graphic  method  for  solving  all  linear  equations  in  one 
unknown  because  every  equation  in  one  unknown  can  be 
thrown  into  a  form  similar  to  4  x  —  4  =  0.  Show  how  this 
may  be  done  with  the  equation  3:r  +  7  =  :r  +  12. 

EXERCISES 

1.  Solve  graphically  the  equation  3x  +  7  =  x  +  12. 

HINT.  The  equation  may  be  written  in  the  form  2  x  —  5  =  0.  Why  ? 

Graph  the  function  2x— 5  just  as  we  graphed  4#  —  4  (Fig.  216). 

See  where  the  graph  of  2  x  —  5  crosses  the  x-axis.  This  is  the 
correct  value  of  x. 

Check  by  substituting  this  value  of  x  in  the  original  equation 
3  x  +  7  =  x  +  12. 

2.  Solve  the  following  linear  equations  by  the  graph,  and 
check  the  results : 

(a)  ox  +  2  =  2x  +  S.  (e)  2.ox  +  9  =  3x  +  7. 

(b)  6x  -  5  =  4x  +  2.  (f)  |_  7  =  x  -  5. 
(c)5a;  +  8  =  8x-4.  x  x 

(d)  11  x  -  9  =  14*  +  7.  ')  3  4      =2~7' 


FUNCTION  INTERPRETED  GRAPHICALLY  305 


HISTORICAL  NOTE.  Perhaps  the  most  important  idea  which  a 
student  can  learn  from  an  elementary  study  of  mathematics  is  the  idea 
of  a  function.  This  is  given  far  greater  prominence  in  European  than 
in  American  texts.  Indeed,  it  is  remarkable  that  though  the  function 
idea  is  generally  admitted  to  be  a  fundamental  notion  in  mathematics, 
the  teaching  of  the  notion  is  often  neglected. 

From  the  simple  illustrations  in  this  text  it  will  appear  that  many 
familiar  facts  and  principles  of  natural  phenomena  can  be  expressed 
as  functions.  A  study  of  these  facts  in  the  form  of  a  mathematical 
function  throws  much  additional  light  on  them.  Thus,  a  clear 
understanding  of  the  principles  of  light,  sound,  and  electricity  could 
not  be  obtained  without  a  study  of  the  mathematical  functions  by 
which  these  principles  are  expressed.  The  dependence  of  one  quantity 
upon  another  is  one  of  the  fundamental  notions  of  human  experience. 
It  is  valuable  to  learn  to  treat  this  notion  mathematically. 

Teachers  of  mathematics  will  also  remember  Professor  Felix  Klein 
as  one  who  has  improved  mathematics  teaching  by  insisting  that  it 
be  made  more  meaningful  to  pupils  of  high-school  age  by  introducing 
and  emphasizing  the  notion  of  a  function. 

345.  Direct  variation.  If  a  man  walks  at  the  rate  of 
3  mi.  an  hour,  the  distance  d  which  he  walks  in  a  given 
time  t  is  found  from  the  equation  d  =  3  t.  The  following 
table  gives  the  distance  passed  over  in  1  hr.,  2  hr.,  3  hr.,  etc. 
and  shows 

1.  That  the  greater  the  time,  the  greater  the  distance 
passed  over. 

2.  That    the   ratio   of   any  time   to  the   corresponding 
distance  is  I. 


Number  of  hours  .  . 

1 

2 

3 

4 

5 

6 

7 

8 

Distance  passed  over 

3 

6 

9 

12 

15 

18 

21 

24 

When  two  numbers  vary  so  that  one  depends  upon  the 
other  for  its  value,  keeping  constant  the  ratio  of  any  value 
of  one  to  the  corresponding  value  of  the  other,  then  one 


306  GENERAL  MATHEMATICS 

is  said  to  vary  directly  as  the  other  or  to  be  rtiwtly  pro- 
portional to  the  other.    Thus  the  number  r  is  said  to  vary 

x 
directly  as  y  if  the  ratio  -  remains  constant,  as  x  and  y  both 

x 
change  or  vary.   The  equation  -  =  k  expresses  algebraically, 

i7 

and  is  equivalent  to,  the  statement  that  JT  varies  directly 

X 

as  y.    The  equation  -  =  k  is  often  written  jc  =  ky.    Show 
why  this  is  correct.    ^ 

EXERCISES 

Translate  the  following  statements  into  equations  of  the 
form      =  /.• : 

y 

1.  The  cost  of  10  yd.  of  dress  goods  is  directly   propor- 
tional to  the  price  per  yard. 

Solution.   I 'sing  c  for  the  total  cost  and  j>  for  the  price  per  yard, 

-  =  10,  or  c  =  10,,. 
P 

This  illustrates  direct  variation,  for  the  greater^the  price  per  yard, 
the  greater  the  total  cost. 

2.  The  railroad  fare  within  a  certain  state  is  3  cents  per 
mile.   Write  the  equation,  showing  that  the  distance  is  directly 
proportional  to  the  mileage. 

3.  The   weight  of  a  mass   of  iron  varies   directly   as   the 
volume. 

4.  If  a  body  moves  at  a  uniform  rate,  the  distance  varies 
directly  as  the  time. 

5.  The  length  (circumference)  of  a  circle  varies  directly  as 
the  diameter. 

6.  The  distance  d  through  which  a  body  falls  from  rest 
varies  directly  as  the  square  of  the  time  t  in  which  it  falls. 
A  body   is   observed  to    fall  400  ft.   in  5  sec.    What  is  the 
constant  ratio  of  d  to  t2  ?    How  far  does  a  body  fall  in  2  sec.  ? 


FUNCTION  INTERPRETED  GRAPHICALLY    307 
Solution.  The  equation  for  d  and  t  is 

j2  =  k.  Why.' 

In  this  problem  *gg-  =  k  ; 

hence  k  =  16. 

Substituting  k  =  16  and  t'2  =  22  in  -  =  k, 


Solving,  f/  =  64. 

This  solution  shows  that  &  may  be  determined  as  soon  as  one 
value  of  t  and  the  corresponding  value  of  d  is  known.  Once  deter- 
mined, this  value  of  k  may  be  used  in  all  problems  of  this  type. 
Thus,  in  all  falling-body  problems  k  =  16  (approx.). 

7.  How  far  does  a  body  fall  from  rest  in  -3  sec.  ?  in  5  sec.  ? 

8.  If  a:  varies  directly  as  y,  and  x  =  40  when  y  =  8,  find 
the  value  of  x  when  y  =  15. 

9.  If  ic  varies  directly  as  a-,  and  -w  =  24  when  x  =  2,  find 
w  when  x  =  11. 

10.  A  stone  fell  from  a  building  576  ft.  high.    In  how1  many 
seconds  did  it  reach  the  ground  ?     (Use  the  method  of  Ex.  6.) 

11.  The  speed  of  a  falling  body  varies  directly  as  the  time. 
Write  the  equation  for  the  speed  V  and  the,  time  t.   A  body 
falling  from  rest  moves  at  the  rate  of  180  ft.  per  second  five 
seconds  after  it  begins  to  fall.    What  will  be  the  speed  attained 
in  nine  seconds  ?. 

12.  The  time  t  (in  seconds)  of  oscillation  of  a  pendulum 
varies   directly  as  the  square  root  of  the   length  I  ;  that  is, 

—-  =  /,-.    A  pendulum  39.2  in.  long  makes  one  oscillation  in 

v/ 

one  second.    Find  the  length  of  a  pendulum  which  makes  an 
oscillation  in  two  seconds. 

13.  The  simple  interest  on  an  investment  varies  directly  as 
the  time.    If  the  interest  for  5  yr.  on  a  sum  of  money  is  $150, 
what  will  be  the  interest  for  6  yr.  4  mo.  ? 


308 


GENERAL  MATHEMATICS 


14.  The  weight  of  a  sphere  of  a  given  material  varies  directly 
as  the  cube  of  its  radius.  Two  spheres  of  the  same  material 
have  radii  of  3  in.  and  2  in.  respectively.  If  the  first  sphere 
weighs  6  lb.,  what  is  the  weight  of  the  second  ? 

346.  Graphing  direct  variation.  Direct  variation  between 
two  quantities  may  be  represented  graphically  by  means 
of  a  straight  line.  Turn  back  to  Chapter  XI,  on  the  formu- 
las, and  find  three  graphs  illustrating  direct  variation. 


-20 


FIG.  217.    GRAPH  OF  C  =  ird  SHOWING  DIRECT  VARIATION 


An  interesting  example  is  furnished   by  graphing  the 

C 
equation  —  =  TT  (where  TT  =  3.14).    This  equation  says  that 

the  circumference  of  a  circle  varies  directly  as  its  diameter. 

Complete  the  following  table,  and  graph  the  results  so 

as  to  obtain  the  graph  in  Fig.  217.    Interpret  the  graph. 


d 

0 

1 

2 

3 

4 

5 

6 

7 

8 

C 

0 

3.1 

6.3 

9.4 

FUNCTION  INTERPRETED  GRAPHICALLY    309 

EXERCISES 

Graph  the  following  examples  of  direct  variation  : 

1 .  v  =  16 1.  (Velocity  equals  16  times  the  number  of  seconds.) 

2.  t  =  5  /.    (Turning  tendency  equals  the  weight  of  5  times 
the  lever  arm.) 

3.  A  =  3b.    (The  area  of  a  rectangle  whose  altitude  is  3 
varies  directly  as  the  base.) 

347.  Inverse  variation.  We  shall  now  consider  a  new 
and  interesting  kind  of  variation.  Suppose  a  gardener 
wishes  to  seed  64  sq.  ft.  of  his  garden  in  lettuce.  If  he 
makes  it  16  ft.  long,  the  width  must  be  4  ft.  (Why?)  If  he 
makes  it  32  ft.  long,  the  width  need  be  only  2  ft.  (Why  ?) 
How  many  possible  shapes  do  you  think  the  gardener 
might  choose  for  his  lettuce  bed  ?  The  following  table 
will  help  you  to  answer  this  question  if  you  remember 
that  it  has  been  decided  that  the  area  shall  be  64  sq.  ft. 


Length  .  .   . 

40 

36 

32 

25 

20 

1(5 

8 

4 

•> 

\ 

Width  .   .   . 

1.6 

1.8 

2 

2.56 

The  table  shows  that  the  length  must  vary  so  as  to  leave 
the  area  constant,  and  that  because  of  this  fact  the 
greater  the  length,  the  smaller  the  width.  The  length  is 
thus  said  to  vary  inversely  as  the  width  or  to  be  inversely 
proportional  to  the  width.  Algebraically  speaking,  a  number 
x  varies  inversely  as  y  if  the  product  xy  remains  constant 
as  both  x  and  y  vary ;  that  is,  if  xy  =  k.  The  student  may 

Jc  k 

also  find  this  equation  written  x  =  -  or  y  =  -• 

y  x 


310  GENERAL  MATHEMATICS 

EXERCISES 

1.  Express  each  of  tin-   following  statements  by   means  of 
equations : 

(a)  The  time  needed  to  go  a  certain  distance  varies  inversely 
as  the  rate  of  travel. 

(b)  The  heat  of  a  stove  varies  inversely  as  the  square  of 
the  distance  from  it. 

(c)  Tke  rate  at  which  a  boy  goes  to  the  corner  drug  store 
varies  inversely  as  the  time  it  takes  him. 

18 

2.  If  .u  =  — ,  show  that  w  varies  inversely  with  z. 

ir 

3.  If  y  varies  inversely  as  x,  and  x  =  12  when  y  =  4,  find 
the  value  of  y  when  x  =  2. 

Solution.    By  definition  of  inverse  variation, 
xy  =  k. 

In  the  first  case,  x  =  12, 

and  y  =  4. 

Therefore  1'J  •  4  =  k, 

or  k  =  48. 

In  the  second  case,  x  =  2. 

Therefore  2  •  y  =  4S,  since  /.•  is  constant. 

Then  y  =  24. 

4.  If  x  varies  inversely  as  ?/,  and  x  =  12  when  y  =  13,  find 
the  value  of  x  when  y  =  2. 

5.  When  gas  in  a  cylinder  is  put  under  pressure,  the  vol- 
ume is  reduced  as  the  pressure  is  increased.     The  physicist 
shows  us  by  experiment  that  the  volume  varies  inversely  as 
the  pressure.    If  the  volume  of  a  gas  is  14  cc.  when  the  pres- 
sure is  9  lb.,  what  is  the  volume  under  a  pressure  of  16  Ib.  ? 

6.  The  number  of  men  doing  a  piece  of  work  varies  inversely 
as  the  time.    If  10  men  can  do  a  piece  of  work  in  33  da.,  in 
how  many  days  can  12  men  do  the  same  piece  of  work  ? 


FUNCTION  INTERPRETED  GRAPHICALLY  311 


40 


•£30 


348.  Graphing  inverse  variation.  We  shall  now  proceed 
to  show  how  inverse  variation  may  be  represented  graph- 
ically. Two  cities  are  48  mi.  apart.  Trains  running  at 
various  rates  carry  the  traffic  between  the  two  cities. 
Suppose  we  attempt 
to  find  out  how 
long  it  will  take  a 
train  which  moves 
uniformly  at  the 
rate  of  24  mi.  per 
hour  to  make  the 
trip,  then  6  mi.  per 
hour,  8  mi.  per  hour, 
etc.  The  following 
table  contains  some 
of  the  values  by 
means  of  which  the 
points  in  Fig.  218 
were  plotted.  The 
equation  represent- 
ing the  situation  is 
48  =  rt.  When  the 
points  of  the  .table 
are  plotted,  it  is  clear  that  they  do  not  lie  on  a  straight 
line,  as  was  the  case  in  direct  variation ;  but  if  they  are 
connected,  the  result  is  the  curved  line  of  Fig.  218.  This 
line  is  one  of  two  branches  of  a  curve  called  a  hyperbola. 


20 


Rate 


FIG.  218.   GRAPH  SHOWING  INVERSE 
VARIATION 


r 

48 

32 

30 

24 

20 

16 

12 

8 

6 

4 

_ 

3 

2 

1 

t 

1 

1.6 

1.6 

2 

312  GENERAL  MATHEMATICS 

EXERCISES 

1.  Determine  from  the  graph  in  Fig.  218  the  time  it  takes 
a  train  whose  rate  is  31  mi.  per  hour ;  20  mi.  per  hour ;  25  mi. 
per  hour. 

2.  Determine  from  the  graph  in  Fig.  218  how  fast  a  train 
runs  which  makes  the  trip  in  1|-  hr.;  2|  hr.;  5^  hr.;  8^  hr. 

3.  Graph  the  equation  ^?y  =  144  (see  Ex.  5,  Art.  347)  and 
interpret  the  graph. 

4.  The  hyperbola  is  an  interesting  mathematical  curve.    See 
if  you  can  help  your  class  learn  more  about  it  by  consulting 
other  books. 

349.  Joint  variation.  In  the  interest  formula  /=  Prt, 
I  depends  for  its  value  on  P,  r,  and  L  A  change  in  any 
one  of  these  letters  causes  a  corresponding  change  in  the 
value  of  /.  We  express  this  by  saying  that  the  interest 
varies  jointly  as  the  principal,  rate,  and  time.  The  alge- 
braic equation  which  defines  joint  variation  is  x  =  kyz. 

EXERCISES 

1.  Turn  to  Chapter  XI,  on  the  formula,  and  find  five  illus- 
trations of  joint  variation. 

2.  If  z  varies  jointly  as  x  and  y,  and  if  «  =  60  when  x  =  3 
and  y  =  2,  find  z  when  x  =  ^  and  y  =  ^. 

*3.  Write  the  following  law  as  a  formula:  The  safe  load  L 
of  a  horizontal  beam  supported  at  both  ends  varies  jointly  as 
the  width  w  and  the  square  of  the  depth  d  and  inversely  as  the 
length  I  between  the  supports. 

*4.  A  beam  12  ft.  long,  4  in.  wide,  and  8  in.  deep  when 
supported  at  both  ends  can  bear  safely  a  maximum  load  of 
1920  Ib.  What  would  be  the  safe  load  for  a  beam  of  the  same 
material  10  ft.  long,  3  in.  wide,  and  6  in.  thick  ? 


FUNCTION  INTERPRETED  GRAPHICALLY    313 

SUMMARY 

350.  This  chapter  has  taught  the  meaning  of  the  fol- 
lowing words  and  phrases :  function,  linear  function,  vari- 
able, dependent  variable,  independent  variable,  constant, 
direct  variation,  inverse  variation,  joint  variation. 

351.  A  clear  understanding  of  the  dependence  of  one 
quantity  upon  another  is  very  important  in  the  everyday 
affairs  of  life. 

352.  A  linear  function  -may  be  treated  algebraically,  or 
geometrically  by  means  of  a  graph. 

3/ 

353.  The  equation  expressing  direct  variation  is  -  =  k. 

tj 

It  is  often  written  x  =  ky.    Its  graph  is  a  straight  line. 

354.  The  equation  expressing  direct  variation  is  xy  =  &, 

k  k 

x  =  -•>   or  y  =  -•     Its   graph   is  a  curved   line  which  is 

y  x 

one  branch  of  a  hyperbola. 

355.  The  equation  expressing  joint  variation  is  x  =  kzy. 


CHAPTER  XIII 

SIMILARITY;  CONSTRUCTION  OF  SIMILAR  TRIANGLES; 
PROPORTION 

356.  Construction  of  similar,  triangles ;  first  method ; 
introductory  exercises.  The  following  exercises  will  help  to 
form  a  basis  for  the  work  of  this  chapter.  The  student 
should  study  them  carefully. 

EXERCISES 

1.  On  squared  paper  lay  off  a  line  segment  AB  of  any  con- 
venient length.    At  .1  construct,  with  the  protractor,  an  angle 
of  32°.    At  B  construct  an  angle  of  63°  and  produce  the  sides 
of  the  two  angles  so  as  to  form  a  triangle.    Call  the  vertex 
angle  C. 

2.  Compare  the  shape  of  the  triangle  ABC  that  you  have 
drawn  with  the  shape  of  those  drawn  by  your  classmates. 

3.  What  was  done  in  Ex.  1  to  insure  that  all  members  of 
the  class  should  get  triangles  of  the  same  shape  ? 

4.  With  the  protractor  measure  angle  (.'  in  your  figure.   How 
else  might  you  have  determined  its  size  ? 

5.  Compare  the  size  of  angle  C  in  your  figure  with  angle  C 
in  the  figures  drawn  by  your  classmates. 

6.  Show  that  any  angle  C  drawn  by  any  member  of  the 
class  ought  to  be  equal  to  any  other  angle  C  drawn. 

7.  Are  the  triangles  drawn  by  the  class  of  the  same  size  ? 
Are  any  two  necessarily  of  the  same  size  '.' 

314 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    315 

357.  Similar  triangles.   Triangles  having  the  same  shape 
are  called  similar  triangles.    Similar  triangles  are  not  neces- 
sarily of  the  same  size.   They  may  be  constructed  by  making 
two  angles  of  one  equal  to  two  angles  of  the  other,  as  was 
done  in  Ex.  1,  Art.  356.    If  two  angles  6f  one  are  equal 
to  two  angles  of  the  other,  it  follows  that  the  third  angles 
are  equal.   (Why  ?)   The  symbol  for  similarity  is  ~->.   Thus 
AABC^AA'B'C"   is  read   "triangle   ABC  is  similar  to 
triangle  A'B'C'"   The  results  of  Art.  356  may  be  summed 
up  in  the  following  geometric  theorem :  If  two  angles  of 
one  triangle  are  equal  respectively  to  two  angles  of  another 
triangle,  the  triangles  are  similar. 

358.  Second  relation  of  parts  in  similar  triangles.    The 
student  should  be  able  to  discover  a  second  method  of  con- 
structing similar  triangles  if  he  studies  and  understands 
the  following  exercises. 

INTRODUCTORY  EXERCISES 

1.  In  the  triangle  ABC  drawn  for  Ex.  1,  Art.  356,  letter  the 
side  opposite  angle  C  with  a  small  letter  c,  the  side  opposite 
angle  B  with  a  small  letter  I,  and  the  side  opposite  angle  A  with 
a  small  letter  a. 

2.  Measure  the  lengths  of  the  sides  a,  l>,  and  c  in  Ex.  1 
to  two  decimal  places.    Find  the  ratio  of  a  to  I,  of  1>  to  c,  of 
a  to  c  (in  each,  case  to  two  decimal  places). 

3.  Compare  your  results  in  Ex.  2  with  the  results  obtained 
by  the  other  members  of  your  class.    What  conclusion  do  you 
make  with  reference  to  the  ratios  of  the  sides  '.' 

359.  Construction  of  similar  triangles ;   second  method. 
The  results  of  Exs.  1-3,  Art.  358,  may  be  summarized  as 
follows :  In  similar  triangles  the  ratios  of  corresponding  sides 
are  equal.    The  work  of  Art.  358  suggests  a  second  method 
for  constructing  similar  triangles. 


3 It)  GENERAL  MATHEMATICS 

EXERCISES 

1.  Draw  a  triangle.     Draw  a  second  triangle  whose  sides 
are    respectively    twice    as    long    as    the    sides    of    the    first 
triangle. 

2.  Compare  the  triangles  constructed  in  Ex.  1  as  to  shape. 
Are  they  similar  ?    Find  the  ratio*  of  the  corresponding  sides. 

3.  Draw  a  triangle  with  sides  three  times  as  long  as  the 
corresponding  sides  of  another  triangle.     Are  they  similar? 
Give  reasons   for  your  answer.     How  do  the  ratios  of  the 
corresponding  sides  compare  ? 

4.  Draw  a  triangle  ABC.    Bisect  the  lines  AB,  AC,  and  EC. 
Call  the  halves  x',  y\  and  z'.    Construct  a  second  triangle,  using 
the  segments  x',  y',  and  z'  as  sides.    Compare  the  two  triangles 
as  to  shape.    Are  they  similar  ?    What  are  the  ratios  of  the 
corresponding  sides  ? 

The  preceding  exercises  suggest  the  following  theorem : 
Two  triangles  are  similar  if  the  ratios  of  the  correspond- 
ing sides  are  equal.  This  gives  us  a  second  method  of 
constructing  similar  triangles ;  namely,  by  making  the 
ratios  of  their  corresponding  sides  equal. 

360.  Construction  of  similar  triangles ;  third  method. 
We  shall  study  a  third  method  of  constructing  similar 
triangles  which  is  suggested  by  the  following  exercises : 

EXERCISES 

1.  Construct  a  triangle  with  two  sides  4.6cm.  and  6.2cm., 
and  with  the  protractor  make  the  included  angle  70°.  Con- 
struct a  second  triangle  with  two  sides  9.2  cm.  and  12.4  cm. 
and  the  included  angle  70°.  Compare  the  triangles  as  to  shape. 
What  is  the  ratio  of  the  corresponding  sides  ?  Measure  the 
corresponding  angles. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    317 


2.  If  convenient  the  class  sholild  divide  itself  into  sections, 
,ne  first  section  constructing  a  triangle  with  two  sides  and  the 
included  angle  as  follows  :  a  =  12,  b  =  18,  and  C  =  40°;  the 
second  section  taking  a  =  8,  b  =  12,  and  C  =  40°;  and  the  third 
section  taking  a  =  4,  b  =  6,  and  C  —  40°.  Compare  the  triangles 
drawn  by  the  three  sections  as  to  shape.  What  is  the  ratio 
of  the  corresponding  sides  ? 

The  preceding  exercises  support  the  geometric  theorem: 
Two  triangles  are,  similar  if  the  ratio  of  two  sides  of  one  equals 
the  ratio  of  two  corresponding  sides  of  the  other,  and  the  angles 
included  between  these  sides  are  equal.  This  theorem  suggests 
the  third  method  of  constructing  similar  triangles. 

361.  Summary  of  constructions  for  similar  triangles.  Two 
triangles  are  similar 

1.  If  two  angles  of  one  triangle  are  constructed  equal  respec- 
tively to  two  angles  of  the  second  triangle. 

2.  If  the  sides  of  the  triangles  are  constructed  so  that  the 
ratios  of  their  corresponding  sides  are  equal. 

3.  If  the  triangles  are  constructed  so  that  the  ratio  of  two 
sides  of  one  is  equal  to  the  ratio  of  two  sides  of  the  other  and 
the  angles  included  between 

these  sides  are  equal. 

362.  Similar  right  tri- 
angles.    We    shall    now 
prove  the  following  the- 
orem:   The  perpendicular 
to  the  hypotenuse  from  the 

vertex  of  a  right  triangle  divides  the  triangle  into  two  triangles 
that  are  similar  to  each  other  (see  Fig.  219). 

Proof.  £x  =  /.x'.  Why? 

Z.V  =  Z/.  Why? 

.-.  £\ADC  ^  AEDC.  Why? 


D 


FIG.  219 


318  GENERAL  MATHEMATICS 

EXERCISES 
(Exs.  1-4  refer  to  Fig.  219) 

1.  Show  that  A.I  IK'  —  A  ABC. 

2.  Show  also  that  A  BCD  ^  A.-lLv '. 

3.  Translate  the  results  of  Exs.  1  and  2  into  a  geometric 
theorem. 

4.  State  a  theorem  expressing  the  results  of  this  article. 

363.  Similar  polygons.  In  later  work  in  mathematics 
we  learn  that  similar  polygons  also  have  corresponding 
angles  equal  and  that  the  ratios  of  the  corresponding  sides 
are  equal.  This  rests  on  the  fact  that  two  similar  polygons 
may  be  divided  into  sets  of  similar  triangles  by  drawing 
corresponding  diagonals  us 
in  Fig.  220. 

Similar  figures  are  of 
frequent  occurrence.  The 
plans  of  construction  work,  FJG  22Q  SIMILAK  POLYGOXS 
drawings  in  shop,  a  sur- 
veyor's copy  of  a  field  triangle,  blue  prints,  a  photograph, 
enlarged  and  reduced  pictures,  are  all  examples.  The  rela- 
tion of  the  different  parts  in  all  the  foregoing  is  shown  by 
magnifying  or  reducing  all  parts  to  a  definite  scale.  Thus, 
you  may  be  able  to  determine  by  looking  at  a  photograph 
of  a  man  that  he  has  large  ears,  although  in  the  picture 
the  actual  measurement  of  either  of  his  ears  may  be  less 
than  a  centimeter.  One  can  tell  by  looking  at  the  plan 
of  a  house  whether  the  windows  are  large  or  small,  be- 
cause the  relation  is  brought  out  by  the  fact  that  all  parts 
are  reduced  to  the  same  scale ;  that  is,  the  ratios  of  the 
corresponding  parts  are  equal.  See  if  you  can  find  examples 
that  will  illustrate  the  last  statement 


CONSTRUCTION  OF  SIMILAR  TKI ANGLES    319 

Similar  triangles  may  be  regarded  as  copies  of  the  same 
triangle  magnified  or  minified  to  a  scale,  or  both  may  be 
regarded  as  scale  drawings  of  the  same  triangle  to  dif- 
ferent scales.  We  shall  study  the  geometric  relations  more 
in  detail  in  the^next  chapter. 

364.  Algebraic  problems  on  similar  figures.  The  fact 
that  the  ratios  of  the  corresponding  sides  of  similar  poly- 
gons are  equal  furnishes  us  with  an  algebraic  method  of 
finding  distances. 

EXERCISES 

1.  In  the  similar  triangles  of  Fig.  221,  if  a  =  3  in.,  a'  =  9  in., 
and  b  =  3  in.,  how  long  is  //  ? 

2.  In    the   similar   triangles    of    Fig.    222,    if   n  =  6  -mm., 
a'  =  8  mm.,   and    I  =  8  mm.,    how   long 

is  b1? 

3.  In  the  similar  triangles  of  Fig.  223,  » 
if  .a' =10.5  mm.,    b  =12  mm.,  and   // =       a/\h 

15  mm.,  how  long  is  a  ? 

Flo    221 

4.  The  sides  of  a  triangle  are  16,  20, 

and  26.    The  shortest  side  of  a  similar 
triangle  is  22.    Find  the  other  sides. 

5.  The  sides  of  a  triangle  are  2.3  cm., 
2.7  cm.,   and   3  cm.     The   corresponding 
sides  of  a  similar  triangle  are  a-,  y,  and 
12cm.    Find  the  values*  of  x  and  y. 

6.  Two  rectangular  boards  are  desired. 
One  is  to  be  4  in.  wide  and  6  in.  long,        . 
the  other  is  to  be  IS^in.  long.    How  wide     «/    \ 


should  the  second  board  be?  FK,  223 

7.  At  a  certain  time  of  day  a  foot  rule 

casts  a  shadow  10  in.  long.   How  long  is  the  shadow  of  a  yard 
stick'at  the  same  time?   Draw  a  diagram  and  prove  your  work. 


820 


( ;  K.\  KK  A  I.   MATHEMATICS 


8.  In  Fig.  224  the  pole,  the  length  of  its  shadow,  and 

the  sun's  rays  passing  over  the  top  of  the  pole  form  a  triangle. 

The  shadow  of  the  pole  is  measured,  and  is  found  to  be  60  ft. 

long.    At  the  same  time  the  shadow  of  a  vertical  stick  2^  ft. 

high  is  measured,  and  is  found  to  be  7|-  ft.  long.  How  may  we 

determine  the  height  of  the  pole 
without  actually  measuring  it  ? 


c 

\ 

\ 

\ 

h 

\ 

\ 

\ 

60' 

FIG.  22 1 


7.5' 
FIG.  225 


Solution.    The   stick?  the    shadow,  and  the   sun's  rays  form  a 

triangle  similar  to  the  first  triangle  (see  Fig.  225).  Why? 
If  we  let  h  denote  the  height  of  the  pole,  we  get 

h        6°  Why? 


Then 


2.5      7.5 

h  =  20. 


z 


9.  The  shadow  of  a  chimney  is  85.2  ft.  long.  At  the  same 
time  the  shadow  of  a  man  6  ft.  2  in.  tall  is  9  ft.  2  in.  How  high 
is  the  chimney  ?  @ 

10.  Draw  a  triangle  ABC  on  squared 
paper  as  in  Fig.  226.   Through  a  point  D  on 
AC  draw  line  DE  II  AB.   Measure  the  seg-    A' 

ments  CD,  DA ,  CE,  and  EB  to  two  decimal 

CD          Cf 

places.  Find  the  ratios and Howdo 

DA         EB 

these  ratios  compare  ?  What  does  this  show  ? 

11.  Draw  a  scalene  triangle  on  squared 
paper,  making  the  base  coincide  with  one 
of  the  horizontal  lines.   Letter  the  triangles 

as  in  Fig.  227.    Choose  any  line  parallel  to  the  base  and  letter 

it  DE  as  in  Fig.  227.    Find  the  ratios  —  —  and  — -7--    How  do 

DA  A  B 

these  ratios  compare?    State  your  conclusion  as  a  theorem. 


FIG.  227 


CONSTR0C-TION  OF  SIMILAR  TUI ANGLES 


n 


FIG.  221 


12.  Suppose  that  in  Fig.  227  Z>C  =2.5  mi.,  DA  =  7.5  mi., 
and  CE  =  9.ini.,  how  long  is  EB  ? 

13.  In  triangle  ABC,  Fig.  227,  the  line  DE  has  been  drawn 
parallel  to  the  base  AB.    Prove  that 

the  small  triangle  CDE  is  similar  to 
the  large  triangle  ABC. 

14.  IiiFig.228,if^U;=3,DE=5, 
and  AB  =  25.5,  how  long  is  BC  ? 

15.  Divide  ;a  line   segment  into 
two  equal  parts  and  show  that  your 
construction  is  correct. 

Construction.  Let  AH,  Fig.  229,  be 
the  line  segment.  Draw  A C  through  A, 
making  any  convenient  angle  with  AB. 
On  AC  lay  off  AD  and  DE,  each 
equal  to  1  unit.  Join  E  to  B.  Draw 
DF II  EB.  Show  that  ,1 F  =  FB. 

HINT,   Use  Ex.  10,  p.  320. 


FIG.  229.    How  TO  BISECT 
A  LINE 


16-.  Divide  a  line  segment  into 
two  parts  whose  ratio  is  |. 

Construction.  Let  AB,  Fig.  230,  be 
the  given  segment.  Draw  A  C,  making 
any  convenient  angle  with  AB,  as 
shown.  On  AC  lay  off  AD  =  2  units 
and  DE  =  3  units.  Join  E  and  B. 
Through  D  draw  DF  II  EB.  Show 
AF  2 


I)' 


Fir,.  230 


that 


FB      3 


17.  Divide  a  line  segment  into 

3    1    2    a 
parts  having  the  ratios-;  ^;  ^;  y 

FIG.  231 

18.  The  distance  across  a  swamp 

•(Fig.  231)  is  to  be  found.    A  point  C  is  located  in  the  same 
line  with  A  and  B.    At  C  and  B  lines  CD  and  BE  are  drawn 


322  GENEKAL  MATHEMATICS 

perpendicular  to  CB,  and  the  line  AD  is  drawn.  The  lengths 
of  CB,  DE,  and  EA  are  found  to  be  80  ft.,  90  ft.,  and  250  ft 
respectively.  Find  the  distance  AB. 

19.  Show  that  the  distance  A  B  across 
the  swamp  could  also  be  found  by  meas- 
uring the  lines  shown  in  Fig.  232. 

365.  Proportion.  The  preceding 
exercises  dealing  with  similar  triangles 
were  solved  by  means  of  a  special 

type  of  equation  expressing  the  fact  that  two  ratios  in  the 
geometric  figure  were  equal.  Thus  in  Fig.  233  the  line  AB 
is  divided  into  two  parts  whose  ratio  is  |  (see  the  method 
of  Ex.  16,  Art.  364).  In  this  construction  it  turns  out  that 
4F  2 

3-        Wh.v'-'     A. f. ,B 

~2*"    ''  '' 

?.  Why?  Vs/ 


DE      3 

AF     AD 
and  »=™-  Why?  Fio.  233 


Such  an  equation,  which  expresses  equality  of  two  ratios, 
is  called  a  proportion.  The  line  segments  AF,  FB,  AD,  and 
DE  are  said  to  be  proportional,  or  in  proportion.  This 
means  that  AF  divided  by  FB  will  always  equal  AD 
divided  by  DE. 

A  proportion  may  thus  be  defined  as  an  equation  which 
expresses  the  equality  of  two  fractions ;  as,  T8:r  =  f.  Another 

ct      c 
example    of    a   proportion    is    -  =  -  •    This   may  be  read 

"  a  divided  by  b  equals  c  divided  by  <?,"  or  "  a  is  to  b  as 
c  is  to  rf,"  or  "  a  over  b  equals  c  over  d."  Sometimes  it  is 
written  a  :  b  =  c :  d,  but  this  form  is  not  desirable. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    323 

EXERCISE 

Is   the   statement  f  =  f    a  proportion  ?   Give  reasons   for 
your  answer.    Is  f  =  -^  a  proportion  ?    Explain  your  answer. 

366.  Means  and  extremes.    The  first  and  last  terms  in 
a  proportion  are  called  the  extremes  and  the  second  and 

third  terms  the  means.   Thus,  in  the  proportion  -  =  —  »  a  and 
d  are  the  extremes  and  b  and  c  the  means. 

EXERCISES 

1.  Compare  the  product  of  the  means  with  the  product  of 
the  extremes  in  the  following  proportions  : 


What  statement  can  you  make  concerning  the  products  ? 

2.  Make  up  several  proportions  and  compare  the  product 
of  the  means  with  the  product  of  the  extremes. 

367.  Theorem  on  the  relation  between  the  means  and  the 
extremes  of  a  proportion.  Exs.  1-2,  Art.  366,  illustrate  a 
well-known  law  or  theorem  ;  namely,  that  in  a  proportion 
the  product  of  the  means  equals  the  product  of  the  extremes. 

a.     c 
If  the  given  proportion  is  -  =  -»  then  the  law  is  algebraically 

stated  thus  :  ad  =  be. 

The  theorem  may  be  proved  as  follows  : 

a      c 

Let  -  =  -  represent 
b      d 

members  by  bd  we  get 


ft  f* 

Let  -  =  -.  represent  any  proportion.     Multiplying  both 
b      d 


abd  _  cbd 

Reducing  each  fraction  to  lowest  terms, 
ad  —  be. 


324  GENERAL  MATHEMATICS 

Since  a  proportion  is  a  special  kind  of  equation,  there 

•  are   special   laws  which    often   make    a  proportion    easier 

to  solve  than  other  equations  which  are  not  proportions. 

The  law  given  on  page  323,  Art.  307,  is  one  of  the  many 

principles  of  proportion  convenient  to  use.    Thus,  instead 

4      16 

of  finding  the  L.C.D.  in  the  equation  -  = — »  and  solving 

o       .  X 

in  that  way,  we  simply  use  the  preceding  law,  and  say 

4  x  =  48. 
x  =  16. 

The  law  is  also  a  convenient  test  of  proportionality, 
since  it  is  usually  simpler  to  find  the  products  than  to 
reduce  the  ratios  to  lowest  terms. 

EXERCISES 

1.  Test    the    following    statements    to    see    if    they    are 
proportions : 

3  _  15  '  .     l5  J  12/  42  w  _^  21  m 

'  7  ~~  35 '  '  2^5  ~~  1.4 '  5  n  ~  2.5  n ' 

5  _  8  2 a  _   5a  11.5  _  7.7 

\~J  ^  —  77'  v*)   o —  ^~K     "  (*•)     o  "    —  o~o * 

<-       11  ox        t  .o  x  o.o        Z.Z 

2.  Find  the  values  of  the  unknowns  in  the  following  pro- 
portions, and  check  by  substituting  in  the  original  equations  : 


(b)  -  .=  ^  •  Solution.    >/-  +  y  -  20  =  yz  —  9. 
•66_l  'y-2»  =  -». 

}  10"5' 

ff-\\  y~12    3  11  +  5   11  -3 

<d>n«iT5'  Check-  irT3=irri 

,   3+1  =  3^  +  2  i->  _  8 

5  14  14"  7' 


CONSTRUCTION, !  (3F  SIMILAR  TRIANGLES    326 

]  a  -  13  =  a  -  14:  ^  ~T~       ~3 

I  — IL — '       .  /i\  x   i    O  _        J- 

'      /T     _I-    .^    ~~    «     _1_     1  U/  .        K' 


3.  If  5  and  3  are  each  added  to  a  certain  number,  and 
1.  and  2  are  each  subtracted  from  it,  the  four  numbers  thus 
obtained  are  in  proportion.    Find  the  number. 

4.  Show  how  to  divide  a. board  54  in.  long  into  two  parts 
whose  ratio  is  y^. 

5.  What  are  the  two  parts  of  a  line  segment  10  cm.  long 
if  it  is  divided  into  two  parts  whose  ratio  is  |  ? 

6.  The  acute  angles  of  a  right  triangle  are  as  2  is  to  5 ; 
that  is,  their  ratio  is  -|.    Find  the  angles. 

7.  If  10°  be  subtracted  from  one  of  two  complementary 
angles  and  added  to  the  other,  the  ratio  of  the  two  angles  thus 
formed  is  ^.    Find  the  angles. 

8.  If  1|  in.  on  a  railroad  map  represents   80  mi.,  what 
distance  is  represented  by  2|  in.? 

9.  Two  books  have  the  same  shape.    One  is  4^  in.  wide; 
and  7-g^  in.  long.    The  other  is  18  in.  long ;  how  wide  as  it  ? 

10.  The  records   of   two  leading   teams   in  the   American 
League  were  Boston  won  68,  lost  32 ;  Chicago  won  64,  lost  36. 
If  the  teams  were  scheduled  to  play  each  other  ten  more  gauit-s. 
how  many  must  Chicago  have  won  to  have  been,  tied  with 
Boston  ? 

11.  If  1  cu.  ft.  of  lime  and  2  cm.  ft.  of  sand  are  used  in 
making  2.4  cu.  ft.  of  mortar,  how  much  of  each  is  needed  to 
make  96  cu.  ft.  of  mortar  ? 

368.  Proportion  involved  in  variation.  Many  problems 
in  physics,  chemistry,  general  science,  domestic  science, 
astronomy,  and  mathematics  may  be  solved  by  either 
variation  or  proportion.  In  fact,  the  whole  theory  of 


826  GENERAL  MATHEMATICS 

proportion  is  involved  in  our  discussion  of  variation,  but 
this  fact  is  not  always  so  obvious  to  a  beginner.  The  fact 
that  problems  may  be  stated  both  in  terms  of  variation  and 
in  terms  of  proportion  makes  it  necessary  for  the  student 
to  recognize  clearly  the  relation  between  variation  and  pro- 
portion. This  relation  will  be  illustrated  in  the  following 
list  of  exercises. 

EXERCISES 

Solve  by  either  variation  or  proportion  : 

1.  If  11  men  can  build  a  cement  walk  in  82  da.,  how  long 
will  it  take  15  men  to  build  it  ? 

(a)  Solution  as  a  variation  problem  : 

mt  =  k.  (The  time   it   takes   to  build 
the  walk  varies  inversely  as 

the  number  of  men.) 
Then  11  •  82  =  k. 

Hence  k  =  902. 

Using  this  value  of  k  in  the  second  case, 

mt  =  902  ; 
but  m  =  15. 

Whence  15  t  =  902, 

and  t  =  *TV  =  60T2S  da. 

(b)  Solution  as  a  proportion  problem.    The  number  of  men  is  not 
in  the  same  ratio  as  the  time  necessary  to  build  the  walk,  but  in 
inverse  ratio  ;  that  is, 


This  proportion  means  "  the  first  group  of  men  is  to  the  second 
group  of  men  as  the  time  it  takes  the  second  group  is  to  the  time  it 
takes  the  first  group." 

Substituting  the  three  known  facts, 

II  -!i 

15  7  82' 

Whence  15  d,  =  902, 

and  c/  =     T*  =  60      da. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    327 

2.  If  200  ft.  of  copper  wire  weighs  60  lb.,  what  is  the  weight 
of  125  ft.  of  the  same  kind  of  wire  ? 

3.  The  cost  of  wire  fencing  of  a  certain  kind  varies  as  the 
number  of  yards  bought.    If  12  rods  cost  $12.80,  how  much 
can  be  bought  for  $44.80? 

4.  Two  men  are  paid  in  proportion  to  the  work  they  do. 
A  can  do  in  24  da.  the  same  work  that  it  takes  B  16  da.  to  do. 
Compare  their  wages. 

5.  A  farmer  has  a  team  of  which  one  horse  weighs  1200  lb. 
and  the  other  1500  lb.    If"  they  pull  in  proportion  to  their 
weight,  where  must  the  farmer  place  the  clevis  on  a  four-foot 
doubletree  so  as  to  distribute  the  load  according  to  the  size 
of  the  horses  ? 

369.  Different  arrangements  of  a  proportion.  The  stu- 
dent will  be  interested  in  seeing  in  how  many  different 
forms  a  proportion  may  be  arranged.  This  he  may  learn 
by  solving  the  exercises  that  follow. 

EXERCISES 

1.  Arrange  the  numbers  3,  6,  7,  and  14  in  as  many  propor- 
tions as  possible.   Do  the  same  for  the  numbers  2,  5,  8,  and  20. 

2.  Can  you  write  two  ratios  that  will  not  be  equal,  using 
the  numbers  2,  5,  8,  and  20  as  terms  of  these  ratios  ? 

3.  How  do  you  decide   which   arrangement   constitutes  a 
proportion  ? 

The  preceding  exercises  suggest  that  a  proportion  such 

as  -  =  -  may  take  four  forms,  as  follows : 
o      a 

(a)  The  given  proportion 


328  GENERAL  MATHEMATICS 

(b)  The  form  obtained  by  alternating  the  means  in  (a): 

a      b 
~c  =  d' 

(c)  The  form  obtained  by  alternating  the  extremes  in  (a): 


(d)  The  form  obtained  by  alternating  both  the  means 
and  extremes  in  (a)  : 

d      b 


The  last  form  can  be  obtained  simply  by  inverting  the 
ratios- in  (a). 

We  know  that  the  proportions  given  above  are  true,  for  by 
applying  the  test  of  proportionality  we  see  that  the  product 
of  the  means  equals  the  product  of  the-  extremes  in  each 
case.  Furthermore,  any  one  of  them  could  have  been 
obtained  by  dividing  the  members  of  the  equation  ad  =  be 

by  the  proper  number.    Thus,  to  get  —  =  -  we  must  divide 
both  members  of  the  equation  ad  =  be  by  ab.    Why  ? 

—.  ad      l<- 

Then  —  •=       : 

ab       ab 

from  which  T  =  -  ,  or  form  (c). 

b      a 

370.  Theorem.  The  preceding  discussion  illustrates  the 
use  of  the  theorem  which  says  that  if  the  product  of  two 
numbers  is  equal  to  the  product  of  two  other  munbers,  either 
pair  may  be  made  the  means  and  the  other  pair  the  extremes 
of  a  proportion. 


CONSTRUCTION  OF  SIMILAR  TKIANGLES    329 

EXERCISES 

1.  Start  with  the  equation  ad  =  be  and  obtain  the  forms 
a       c     a       b  d       b 

7  =  3'  -  =  3  '  und  -  =  -  •    This  completes  the  proof  of  the 
(>       a     c       a  c      a 

theorem  just  stated.    Why  ? 

2.  Write  the  four  possible  forms  that  can  be  obtained  from 
the  following  products  : 

(a)  5  •  21  =  7  •  15.  (o)  3  a  .  4  b  =  a  •  12  b. 

(b)  3  •  28  =  12  •  7.  (d)  15  -  7  t  =  105  /.  .  ( 


=      .  she,  that      =. 


FIG.  234 

4.  If  two  equilateral  polygons  have  the  same  number  of 
sides,  the  corresponding  sides  are  proportional  (see  Fig.  235). 

i  n 

Proof.  =1,  Why? 

JJ  o 

and  1- 


TIV  .  A,r,     .> 

Therefore  —  ;  =  —  —  •  \v  hy  ? 

J5(        />  C 

By  alternating  the  means, 

,4£        BC 


A'E'      B'C' 
and  so  on  for  the  other  corresponding  sides. 

371.  Mean  proportional.     In  the  proportion  -=  -,  b  is 

called  the  mean  proportional  between  a  and  c  (note  that 
b  appears  twice  in  the  proportion). 


330  GENERAL  MATHEMATICS 

EXERCISES 

1.  What  is  a  mean  proportional  between  4  and  9  ? 
HINT.   Let  x  =  the  number. 

TU  *  X 

Ineii  -  =  -. 

x      y 

From  which  x-  =  36. 

Then  r  =  ±  6. 

2.  Show  that  the  value  of  b  in  the  proportion  -  =  -  is  given 

0          C 

by  the  equation  b  =  ±  Vac  (read  "  &  equals  -f-  or  —  the  square 
root  of  ac"~). 

3.  "What  is  a  mean  proportional  between  2  and  18  ?  between 
10  and  40  ?  between  2  and  800  ? 

4.  Find  a  mean  proportional  between  a?  and  i2;  between 
3?  and  y3. 

372.  How  to  pick  out  corresponding  sides  of  similar 
triangles.  The  similar  triangles  of  Fig.  236  are -placed  so 
that  in  certain  cases  a  line  is  a  side  in  each  of  two  similar 
triangles.  Thus,  AC  is  a 
side  of  A  ADC  and  also  of 
the  similar  triangle  ABC. 
This  suggests  that  the 
line  may  occur  twice  in 
the  proportion  of  the  cor- 
responding  sides.  In  this  FlG  236 

way  it  is  seen  that  the 
line  becomes  a  mean  proportional  between  the  other  two. 

This  analysis  can  be  checked  only  by  actually  writing 
the  proportion  of  pairs  of  corresponding  sides  of  similar 
triangles.  In  order  to  do  this  correctly  the  student  must 
remember  that  £he  (•nrrexpondinri  sides  of  similar  triangles 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    381 

are  the  sides  which  lie  opposite  equal  angles.  Hence,  from 
the  fact  that  A  ADC  >—  A  ABC  we  may  write  the  following 
proportion : 

AD  (opposite  Z  z  in  A  ADC)  _  ^4  C  (opposite  Z  2:  in  A  ADC) 
A C (opposite  Z  z'  in  A  AB (7)      ^(opposite  Z.C  in  AACB) 

™,  AD      AC 

1  hat  is,  = 

AC      AB 

AC  is  thus  seen  to  be  the  mean  proportional  between  AD 
and  AB. 

Show  in  a  similar  way  that  BC  is  a  mean  proportional 
between  the  hypotenuse  AB  and  the  adjacent  segment  BD ; 

BD      BC   . 

that  is,  show  that  - —  =  — -• 
BC      AB 

We  may  summarize  the  preceding  exercises  and  dis- 
cussion by  the  theorem:  In  a  right  triangle  either  side 
including  the  right  angle  is  a  mean  proportional  between  the 
hypotenuse  and  the  adjacent  segment  of  the  hypotenuse  made 
by  a  perpendicular  from  the  vertex  of  the  right  angle  to  the 
hypotenuse. 

373.  Theorem.  If  in  a  right  triangle  a  perpendicular  is 
drawn  from  the  vertex  of  the  right  angle  on  the  hypotenuse, 
the  perpendicular  is  a  mean  proportional 
between  the  segments  of  the  hypotenuse. 
The  truth  of  the  preceding  theorem  will 
be  seen  from  the  following:  A  D 

In  AABC  (Fig.  237)  Z  C  is  a  right  Fio.  237 

angle,  and  CD  ±  AB.  |f  =  ff  because  A  ADC  -  A  CDS, 
and  the  corresponding  sides  are  therefore  in  proportion. 


332    ;  :...-.    ..GENERAL  MATHEMATICS-,    ...;_ 

EXERCISES 

1.  Write  out  the  complete  proof  for  the  preceding  theorem. 

2.  Find  the  altitude  drawn  to  the  hypotenuse  of  a  right 
triangle  if  it  divides  the  hypotenuse  into  two  segments  whose 
lengths  are  4  in.  and  16  in.  respectively.    Find  also  each  leg 
of  the  right  triangle. 

*3.  In  a  right  triangle  ABC  (right-angled  at  C)  a  perpen- 
dicular is  drawn  from  C  to  AB.  If  CD  =  8,  then  A  I)  =  4. 
Find  the  length  of  AB. 

374.   Construction  of  a  mean  proportional.    The  theorem 
of  Art.  ^73  on  page  331  furnishes  us  with  a  method  of  con- 
structing a  mean  proportional  between  any          |       b       [ 
two  line  segments,  as  will  now  be  shown. 

In  Fig.  238  we  are  given  two  line  seg-    '  -  ' 
ments  a  and  b.     The  problem  is  to  con- 
struct a  mean  proportional  (say  x  units  long)  between  a  and  b. 

CL         CC 

We  know  that  the   equation  -  =  -  will  represent  the 

x      b 
situation, 

Construction.    On  a  working  line,  as  A  K  in  Fig.  239,  we  lay  off  a 
from  A    to   B   and  b  from  B  to   C:    With  AC  as  a  diameter  we 
construct   a  semicircle.    At 
B  we  erect  a  perpendicular  ^--       ~~~ 


^ 


B  OK 


intersecting  the  circle  at  D.  /' 

Then  BD   is    the    required 

mean  proportional.  '       /' 

I    /' 
Proof.   Connect  A  with  D        [^_  _  rJL 

and  C  with  D.    Then  BD  is        l 

the   required    mean  propor-  FIG.  239.    MEAN  PROPORTIONAL 

tional  between  a  and  b  pro-  CONSTRUCTION 

vided  we  can  show  that  Z-D 

is  a  right  angle.    (Why?)   We  shall  proceed  to  show  that  ZD  is  a 

right  angle  by  proving  that  if  any  point  on  a  circle  is  connected  with 

the  ends  of  a  diameter,  the  angle  formed  at  that  point  is  a  right  angle. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    333 


In  Fig.  240  we  have  a  given  circle  constructed  on  the  diameter 
AC  and  a  point  D  connected  with  the  ends  of  AC.  We  must 
show  that  ZZ)  is  a  right  angle. 

Connect  D  and  0. 

Then     Z  z  =  Z  .s  +  Z  .s',          (1) 

because  an  exterior  angle  of  a  tri- 
angle is  equal  to  the  sum  of  the 
two  nonadjacent  interior  angles; 


and 


Z  y  =  Z  t  +  Z  I' 


(2) 


for  the  same  reason. 

By  adding  (1)  and  (2), 


FIG.  240 


Since 

But 

and 

Therefore 

Then 


Z  x  +  Z  y  =  Z  ,s-  +  Z  /  +  Z  t  +  Z  ('. 
Z  x  +  Z  y  -  180°, 


Z  <  =  Z  <'. 

2  Z.v  +  2  Z<  =  180°. 
Zs  +  Z<  =  90°. 
ZZ)--90°. 


Why? 
Why  ? 

Why? 

..Why? 

Why? 

Why? 


Then,  if  in  Fig.  239  Z  D  =  90°,  the  proportion  -  —  is  true, 


and  BD  is  a  mean  proportional  between  a  and 


BD        b 
Give  reasons. 


EXERCISES 


1.  Explain  how  a  mean  proportional  between  two  given 
line  segments   may  be  constructed. 

2.  Construct  a  mean  proportional  be- 
tween 9  and  16,  4  and  16,  4  and  9,  16 
and  25,  25  and  36.  A 

3.  In   triangle  ABC,  Fig.   241,  Z.C 

is  a  right  angle,  CDA.AB,  AD  =  2,  and  DB  =  6.    Find  the 
lengths  of  AC  and  CB. 


334  GENERAL  MATHEMATICS 

4.  Find  the  mean  proportional  between  the  line  segments 
m  and  n  in  Fig.  242. 

5.  Measure  in  and  n  in  Fig.  242  and  the  mean  proportional 
constructed  in  Ex.  4.    Square  the  value  of  m 

the  mean  proportional    and   see   how  the 

value   compares    with   the    product   of    m    , n | 

and   n-  Fir..  242 

*6.  Construct  a  square  equal  in  area  to  a 

given  rectangle ;  to  a  given  parallelogram  ;  to  a  given  triangle. 

a      c 
375.  Fourth  proportional.    In  the  proportion  T  =  -^'  d  is 

0  \Jv 

called  the  fourth  proportional  to  «,  6,  and  c.  There  are 
two  methods  of  finding  the  fourth  proportional  to  three 
given  numbers  a,  5,  and  c. 

Algebraic   method.    Let    x    represent    the    value    of   the   fourth 
proportional. 

TI,  a      c 

Then  -  =  - 

ft      x 

(by  definition  of  a  fourth  proportional). 
Solving  for  x,  ax  =  be. 
be 


x_ A  c         F  G      V 

a 

FIG.  243.   How  TO  CONSTRUCT 

Geometric  method.  Take  three  given         A  FOURTH  PROPORTIONAL 
lines,  as  a,  b,  and  c   in  Fig.  243,  and 

draw  any  convenient  angle,  as  Z.BAC.  On  AB  lay  off  AD  =  a, 
DE-l.  On  the  other  line  A  Clay  oSAF=c.  Draw  DF.  Then  draw 
EG  II  DF  as  shown.  Then  FG  is  the  required  fourth  proportional. 

See  if  you  can  show  why  the  construction  is  correct. 

EXERCISES 

1.  Check   the   construction  above  by  measuring  the  four 

a       c 

segments  to  see  if  —  =  -  • 
o      x 

2.  Construct  a  fourth  proportional  to  three  given  line  seg- 
ments 2  cm.,  3  cm.,  and  5  cm.  long  respectively. 


CONSTRUCTION  OF  SIMILAR  TKIA^ULES    33o 


3.  Show  algebraically  that  the  segment  obtained  in  Ex.  '1 
should  be  1\  cm.  long. 

4.  Construct  a  fourth  proportional  to  three  segments  5  cm., 
6  cm.,  and  9  cm.  long  respectively. 

5.  Check  your  work  in  Ex.  4  by  an  algebraic  method. 

*  376.  To  find  the  quotient  of  two  arithmetical  numbers  by 
a  special  method.   To  find  ||  in  per  cent  we  need  to  solve  the 

22        x 
equation     ^  =  w 

(Why?)    This    pro- 

portion suggests  simi- 

lar triangles.    If  we 

take  a  horizontal  line 

OM    (Fig.  244)    as 

a    dividend    line    on 

squared    paper,    and 

ON  perpendicular-  to 

OM  as  a  divisor  line, 

then  lay  off   OA  on 

OM  equal  to  22  units, 

and  at  A  erect  a  per- 

pendicular and  mark 

off  AB  equal  to  70 

units,  we   can   solve 

our  problem  provided 

we  draw  another  line  DR  100  units  above  OM  and  parallel 

to  it.    Call  DR  the  quotient  line. 

Stretch  a  string  fastened  at  0  so  that  it  passes  through 
J5,  meeting  the  quotient  line  at  C. 


Di 


de 


nd  -Li 


-M- 


FIG.  244 


Then 


-      or 


NOTE.    The  proof  is  left  to  the  student. 
Therefore  22  is  approximately  31%  of  70. 


330  ( J  KN  K1J  A  L   M  ATH.EMATK  'S 

EXERCISES 

*1.  Point  out  the  similar  triangles  in  the  device  for  express- 
ing quotients  "used  in  Fig.  244.  Read  the  sides  which  are 
proportional. 

*2.  A  gardener  planted  12  A.  of  potatoes,  8  A.  of  beans, 
13  A.  of  onions,  3  A.  of  celery,  and  5  A.  of  cabbage.  By  means 
of  the  device  used  in  Fig.  244  show  the  distribution  of  his 
garden  in  per  cents. 

377.  Verbal  problems  solved  by  proportion.  We  have 
said  that  many  problems  of  science,  the  shop,  and  engineer- 
ing can  be  solved  by  proportion.  We  shall  proceed  to 
study  how  to  solve  some  of  these  problems  by  using  our 
knowledge  of  proportion. 

In  the  study  of  turning  tendency,  Art.  233,  we  recognized 
the  following  familiar  principle  of  the  balanced  beam :  The 
left  weight  times  the  left  lever  arm  equals  the  right  weight  times 
the  right  lever  arm.  As  a  formula 

•    I~-  7-77  A  F  B 

tins  may  be  written  wlll  =  u'2lz.  i      .     . 


100 
Ib. 


60 
Ib. 


This  principle  is  already  familiar 
to  all  who  have  played  with  a 
seesaw.  They  discovered  long  ago  FiG  245 

that  a  teeter  board   will    balance 

when"  equal  products  are  obtained  by  multiplying  the 
weight  of  each  person  by  his  distance  from  the  point  of 
support  (fulcrum). 

If,  in  Fig.  245,  B  weighs  60  Ib.  and  is  5  ft.  from  the  ful- 
crum F,  then  A,  who  weighs  100  Ib.,  must  be  3  ft.  from 
the  fulcrum.  Thus,  60  •  5  =  100  •  3  is  a  special  case  of 
general  law  wll1  =  w%lz. 

If  we  divide  both  members  of  the  equation  iv^^  =  W212  by 

7  nn 

w^ly  we  get  -  =  — - ,  which  is  in  the  form  of  a  proportion. 
'2      wi 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    337 


The  student  will  learn  that  in  shop  work  many  problems 
dealing  with  the  lever  or  balanced  beam  may  be  solved  by 
some  form  of  the  preceding  proportion. 

BEAM  PROBLEMS 

1.  A  lever  (Fig.  246)  10  ft.  long  carries  weights  of  40  Ib. 
and  50  Ib.  at  its  ends.  Where  should  the  fulcrum  be  placed 
so  as  to  make  the  lever  balance  ? 

Solution.    Let  x  =  the  number  of  feet  from  F  to  A , 

and  10  —  x  =  the  number  of  feet  from  /•'  to  B. 

Then       50 x  =  40(10  -x).    Why?        A       x        F       10-x       B 
50x  =  400-40x.  A  A 


90  x  =  400. 


50 

Ib. 


\40\ 
lib. 


FIG.  246 


2.  A,  who  weighs  80  Ib.,  sits  6  ft.  from  the  fulcrum.    If  B 
weighs  100  Ib.,  at  what  distance  from  the  fulcrum  should  A 
sit  in  order  to  balance  B  ? 

3.  A  and  B  together  weigh  220  Ib.    They  balance  when  A  is 
5  ft.  and  B  6  ft.  from  the  fujcrum.    Find  the  weight  of  each. 

4.  A  and  B  are  4  ft.  and  6  ft.  respectively  from  the  fulcrum. 
If  B  weighs  60  Ib., 

how  much  does  A 
weigh  ? 

5.  How      could 
you  weigh  yourself 
without  a  scale  ? 

6.  AB  in  Fig.  247 
is  a  crowbar  8|  ft. 

long  supported  at  F,  \  ft.  from  A .  A  stone  presses  down  at 
A  with  a  force  of  2400  Ib.  How  many  pounds  of  force  must 
be  exerted  by  a  man  pressing  down  at  />'  to  raise  the  stone  ? 
(Disregard  the  weight  of  the  crowbar.) 


FIG.  247 


GESEKAL  MATHEMATK  > 


7.  In  attempting  to  raise  an  automobile  (Fig.  248)  a  man  lifts 
with  a  force  of  150  Ib.  at  the  end  of  a  lever  10  ft.  long.  The  distance 
from  the  axle  to  F  is  2£  ft. 

What  force  is  exerted  up- 
ward on  the  axle  as  a  result 
of  the  man's  lifting  ? 

8.  Find  ^  if  19  =  18  ft., 
>/-2  =  62  Ib.,  and  w1  =  51  Ib. 

9.  Find  Ja  if    ^  =  40  in., 
»/-2  =26  Ib.,  and  ^=38  Ib. 


FIG.  248 


MIXTURE  AND  ALLOY  PROBLEMS 

1.  How  much  water  must  be  added  to  10  gal.  of  milk, 
testing  §\°fo  butter  fat,  to  make  it  test  4%  butter  fat? 
Solution.    Let  x  =  the  number  of  gallons  of  water  added. 

Then       x  +  10  =  the  number  of  gallons  of  diluted  milk, 

51 
and  — —  •  10  =  the  amount  of  butter  fat  in  the  undiluted  milk. 

I  $ff  (x  +  10)  =  the  amount  of  butter  fat  in  the  diluted  milk. 
Since  the  amount  of  butter  fat  remains  constant, 

-$1    10=     *    (x 
100 

110       a: +  10 
200 


Why  'i 
Whv  '{ 


25 
4  x  +  40  =  fjo. 

x  =  3  1,  the  number  of  gallons  of  water  to  be  added. 

2.  A  physician  has  a  25%  mixture  of  listerine  in  water.   How 
much  water  must  he  add  to  it  to  make  it  a  15      mixture  ? 


Solution.    Consider   an    arbitrary  quantity  of  the  mixture,  say 
100  oz. 

Let  x  =  the  number  of  ounces  of  water  added  to  every 

100  oz.  of  the  mixture. 
Then     100  +  x  =  the  number  of  ounces  in  the  new  mixture. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    339 

Since  25%  of  the  original  mixture  is  listerine, 

25 
— — =  the  per  cent  of  listerine  in  the 

new  mixture. 
And  since  15%  of  the  new  mixture  is  to  be  listerine, 

25       =JL5_ 
100  +  x      100 ' 

1500  +15  a:  =  2500. 
15  x  =  1000. 

Hence  66$  oz.  of  water  must  be  added  to  every  100  oz.  of  the 
original  mixture. 

3.  How  much  water  should  be  added  to  a  bottle  contain- 
ing 4oz.  of  the  original  mixture  in  Ex.  2  to  make  it  a  15% 
mixture  ? 

4.  If  a  patent  medicine  contains  30%  alcohol,  how  much 
of  other  ingredients  must  be  added  to  12  qt.  of  it  so  that 
the  mixture  shall  contain  only  20%  alcohol  ? 

5.  How  many  quarts  of  water  must  be  mixed  with  30  qt. 
of  alcohol  82  %  pure  to  make  a  mixture  70  %  pure  ? 

6.  "\Yhat  per  cent  of  evaporation  must   take   place   from 
a  5%   solution  of  salt  and  water  (of  which   5%   by  weight 
is  salt)  to  make  the  remaining  portion  of  the  mixture  a  7% 
solution  ? 

7.  Two  grades  of  coffee  costing  a  dealer  25$  and  30$  per 
pound  are  to  be  mixed  so  that  50  Ib.  of  the  mixture  will  be 
worth   28$  per  pound.    How  many  pounds  of  each  kind  of 
coffee  must  be  used  in  the  mixture  ? 

8.  In  a  mass  of  alloy  for  watch  cases  which  contains  80  oz. 
there  are  30  oz.  of  gold.    How  much  copper  must  be  added 
in  order  that  in   a  case  weighing  2  oz.  there  shall  be  -}»>/. 
of  gold  ? 


340  GENERAL  MATHEMATICS 

Solution.    Let       x  =  the  number  of  ounces  of  copper  to  be  added. 
Then  80  4-  x  =  the  number  of  ounces  in  the  new  alloy. 

-  =  the  ratio  between  the  whole  mass  of  alloy 
30 

and  the  gold. 

y  =  the  ratio  of  a  sample  of  the  new  alloy  to  the 
gold  in  the  sample. 

ri-,1  80    +  X  \ 

Then  —_  =  _..  Why? 

80  +  z  =  120.  Why? 


Hence  40  oz.  of  copper  should  be  added. 

9.  In  an  alloy  of  gold  and  silver  weighing  80  oz.  there  are 
10  oz.  of  gold.  How  much  silver  should  be  added  in  order  that 
10  oz.  of  the  new  alloy  shall  contain  only  ^  oz.  of  gold  ? 

10.  Gun  metal  is  composed  of  tin  and  copper.  An  alloy  of 
2050  Ib.  of  gun  metal  of  a  certain  grade  contains  1722  Ib.  of 
copper.  How  much  tin  must  be  added  so  that  1050  Ib.  of  the 
gun  metal  may  contain  861  Ib.  of  copper  ? 

*378.  Specific-gravity  problems.  A  cubic  foot  of  glass 
weighs  2.89  times  as  much  as  a  cubic  foot  of  water  (a 
cubic  foot  of  water  weighs  62.4  Ib.).  The  number  2.89  is 
called  the  specific  gravity  of  glass.  In  general,  the  specific 
gravity  of  a  substance  is  defined  as  the  ratio  of  the  weight 
of  a  given  volume  of  the  substance  to  the  weight  of  an  equal 
volume  of  water  at  4°  centigrade.  What  would  it  mean, 
therefore,  to  say  that  the  specific  gravity  of  14-karat  gold 
is  14.88  ?  A  cubic  centimeter  of  distilled  water  at  4°  cen- 
tigrade weighs  just  1  gm.  Since  the  specific  gravity  of 
14-karat  gold  is  14.88,  one  cubic  centimeter  of  gold  weighs 
14.88  gm.,  2  cc.  weighs  29.76  gm.,  etc.  In  short,  the  weight 
of  an  object  in  grams  equals  the  product  of  its  volume  in 
centimeters  times  its  specific  gravity. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    341 

EXERCISES 

1.  How  many  cubic  centimeters  of  distilled  water  (specific 
gravity  equal  to  1)  must  be  mixed  with  400  cc.  of  alcohol 
(specific  gravity  equal  to  0.79)  so  that  the  specific  gravity  of 
the  mixture  shall  be  0.9? 

HINT.  Find  the  weight  of  the  two  parts  and  set  the  sum  equal 
to  the  weight  of  the  mixture. 

2.  Would  you  accept  half  a  cubic  foot  of  gold  on  the  con- 
dition that  you  carry  it  to  the  bank  ?    Explain  your  answer. 

3.  Brass  is  made  of  copper  and  zinc.  Its  specific  gravity  is  8.5. 
How  many  cubic  centimeters  of  copper  (specific  gravity  8.9)  must 
be  used  with  100  cc.  of  zinc  (specific  gravity  7.15)  to  make  brass  ? 

4.  What  is  the  specific  gravity  of  a  steel  sphere  of  radius 
1  cm.  and  weight  32.7  gm.  ? 

379.  Proportionality  of  areas.  The  geometric  exercises 
to  be  given  in  this  article  are  important.  The  student 
should  study  them  carefully,  prove  them,  and  try  to 

illustrate  each. 

EXERCISES 

1.  Prove  that  the  areas  of  two  rectangles  are  to  each  other 
as  the  products  of  their  correspond- 
ing dimensions. 

Proof.    Denote  the  areas  of  the  rec- 
tangles by  #j  and  7?2,  as  in  Fig.  249,     

and  their  dimensions  as  shown.  ' 

Rl  =  albl.  Why? 

Rz  =  a2b2.  Why? 

Therefore      5l  =  2LI.  Why? 

R2      a  A  FIG.  249 

It  is  important  to  note  that  the  last  proportion  is  obtained  by 
dividing  the  members  of  the  first  equation  by  those  of  the  second. 


842  GENERAL  MATHEMATICS 

2.  If  two  rectangles  (Fig.  250)  have  equal  bases,  they  are 
to  each  other  as  their  altitudes.    (Follow  the  method  of  Ex.  1.) 

3.  If  two  rectangles    have  equal  alti- 
tudes, they  are  to  each  other  as  their  bases. 

4.  The  area  of  a  rectangle  is  48  sq.  ft. 
and  the  base  is  11  yd.    What  is  the  area  of 
a  rectangle  having  the  same  altitude  and 
a  base  equal  to  27.5  yd.  ? 

5.  .Prove  that  the  areas  of  two  parallelo- 
grams are  to  each  other  as  the  products  of 
their  bases  and  altitudes. 

6.  The  areas   of  two  triangles  are  to 

each   other   as   the    products    of   their   bases    and    altitudes. 

7.  The  areas  of  two  parallelograms  having  equal  bases  are 
to  each  other  as  their  altitudes  ;  the  areas  of  two  parallelograms 
having  equal  altitudes  are  to  each  other  as  their  bases. 

8.  The  areas  of  two  triangles  having  equal  bases  are  to 
each  other  as  their  altitudes  ;  the  areas  of  two  triangles  having 
equal  altitudes  are  to  each  other  as  their  bases. 

9.  Prove  that  triangles  having  equal  bases  and  equal  alti- 
tudes are  equal. 

*10.  Construct  the  following  by  means  of  Ex.  9:  a  right  tri- 
angle, an  isosceles  triangle,  an  obtuse-angled  triangle,  each  equal 
to  a  given  triangle. 

SUMMARY 

380.  This  chapter  has  taught  the  meaning  of  the  fol- 
lowing   words     and    phrases :     similar    triangles,    similar 
polygons,    proportion,    means,    extremes,    fulcrum,    mean 
proportional,  fourth  proportional,  alloy,  specific  gravity. 

381.  Polygons  that  have  the  same  shape  are  similar. 

382.  In  similar  triangles  the  corresponding  angles  are 
equal  and  the  corresponding  sides  are  in  proportion. 


CONSTRUCTION  OF  SIMILAR  TRIANGLES    343 

383.  Two  similar  triangles  may  be  constructed  by 

1.  Making  two  angles  of  one  equal  to  two.  angles  of 
the  other. 

2.  Making  the  ratios  of  corresponding  sides  equal. 

3.  Making  the  ratio  of  two  sides  of  one  equal  to  the 
ratio  of  two  sides  of  the  other,  and  the  angles  included 
between  these  sides  equal. 

384.  A  proportion  expresses  the  equality  of  two  ratios. 

385.  A  convenient  test  of  proportionality  is  the  theorem 
that  says  the  product  of  the  means  equals  the  product  of 
the  extremes. 

386.  If    ad  =  be,    we    may    write    the    following    four 
proportions  : 

-       a      c  a      b  d      c  ...    d      b 


387.  The  fact  that  the  ratios  of  corresponding  sides  of 
similar  polygons  are  equal  furnishes  us  with  an  algebraic 
method  of  finding  distances. 

388.  Inaccessible  distances  out  of  doors  may  often  be 
determined  by  means  of  a  proportion. 

389.  Beam  problems  and  mixture,  alloy,  and  specific- 
gravity  problems  may  be  solved  by  equations  which  take 
the  form  of  proportions. 

390.  If  a  line  is  drawn  parallel  to  the  base  of  a  triangle, 
the  triangle  cut  off  is  similar  to  the  given  triangle,  and  the 
corresponding  sides  are  in  proportion. 

391.  The  following  important  theorems  about  the  area 
of  two  parallelograms  have  been  proved  : 

1.  The  areas  of  two  parallelograms  are  to  each  other 
as  the  product  of  their  bases  and  altitudes. 


344  GENERAL  MATHEMATICS 

2.  The  areas  of  two  parallelograms  having  equal  bases 
are  to  each  other  as  their  altitudes,  and  the  areas  of  two 
parallelograms  having  equal  altitudes  are  to  each  other  as 
their  bases. 

392.  Three  theorems  similar  to  those  in  Art.  391  were 
proved  for  the  areas  of  rectangles  and  triangles. 

393.  If   in  a  right   triangle  a  line  is  drawn  from  the 
vertex  of  the  right  angle  perpendicular  to  the  hypotenuse, 

1.  The  triangle   is  divided    into    two    similar    triangles 
which  are  each  also  similar  to  the  given  triangle. 

2.  The  perpendicular  is  a  mean   proportional  between 
the  two  segments  of  the  hypotenuse. 

3.  Either  side  about  the  right  angle  is  a  mean  propor- 
tional   between   the  whole  hypotenuse   and   the   adjacent 
segment. 

394.  The  following  constructions  have  been  taught: 

1.  How  to  construct  a  mean  proportional. 

2.  How   to   construct  a  square   equal  to  a  given  rec- 
tangle, parallelogram,  or  triangle. 

3.  How  to   construct  a  right  triangle   or  an   isosceles 
triangle  equal  to  a  given  scalene  triangle. 

4.  How  to  construct  a  fourth  proportional. 

5.  How  to  divide  a  line  segment  into  two  parts  which 
have  a  given  ratio. 


CHAPTER  XIV 

INDIRECT  MEASUREMENT;  SCALE  DRAWINGS; 
TRIGONOMETRY 

395.  Scale  drawings.    Up  to  this  point  we  have  made 
several  uses  of  scale  drawings.   In  Chapter  X  the  relation  of 


FIG.  251.    SCALK  DRAWING  OF  A  LIBRARY  TABLE 
(Courtesy  of  Industrial  Arts  Magazine) 

quantities  was  shown  by  line  segments  whose  proportional 
lengths  represented  the    relative    size    of   the   quantities. 


345 


34l>  GEXE11AL  MATHEMATICS 

In  another  form  of  graphic  work,  scale  drawings  have 
helped  us  to  understand  the  meaning  of  functions,  equa- 
tions, and  formulas.  In  addition  to  the  foregoing,  scale 
drawings  are  probably  familiar  to  the  student  in  the  form 
of  shop  drawings,  geography  maps,  blue  prints,  maps  in 
railroad  guides,  and  architects'  plans. 

The  shop  drawing  in  Fig.  251  illustrates  a  use  of  a  scale 
drawing,  which  we  shall  now  study  in  some  detail. 

The  figure  shows  that  a  scale  drawing  gives  us  an  accu- 
rate picture  of  the  real  object  by  presenting  all  the  parts 
in  the  same  order  of  arrangement  and  showing  the  relative 
sizes  graphically  by  means  of  proportional  line  segments. 
Obviously  this  fact  rests  on  the  principle  of  similarity,  and 
the  ratio  between  any  two  line  segments  in  the  plan  equals 
the  ratio  between  the  lengths  of  the  two  corresponding 
parts  of  the  library  table  (Fig.  252). 

By  means  of  the  scale  drawing  we  are  able  to  deter- 
mine the  dimensions  of  parts  of  the  table  even  though 
they  are  not  given  on  the  plan.  In  fact,  in  the  case  of 
architects'  and  surveyors'  scale  drawings  we  are  able  to 
measure  lines  which  in  the  real  object  are  inaccessible. 

This  last  procedure  illustrates  precisely  the  use  which 
we  want  to  make  of  scale  drawings  in  this  chapter.  In 
many  cases  we  shall  want  to  measure  distances  that  can- 
not be  measured  directly  with  steel  tape  or  other  surveying 
devices;  for  example,  (1)  the  heights  of  towers,  buildings, 
or  trees ;  (2)  the  width  of  ponds,  lakes,  or  rivers  ;  (3)  the 
length  of  boundary  lines  passing  through  houses,  barns,  or 
other  obstructions. 

We  can  usually  determine  such  distances  by  following 
the  method  set  forth  in  the  following  outline: 

1.  Measure  enough  actual  lines  and  angles  in  the  real 
object  so  that  a  scale  drawing  of  the  object  can  be  made. 


TRIGONOMETRY 


347 


2.  Draw  the  figure  to  scale,  preferably  on  squared  paper. 

3.  Measure  carefully  with  the  compasses  and  squared 
paper  the  lines  in  the  figure  which  represent  the  inaccess- 
ible lines  of  the  actual  object  that  is  being  considered. 


FIG.  252.   THE  FINISHED  LIBRARY  TABLE 
(Courtesy  of  Industrial  Arts  Magazine) 

4.  Translate  the  measurements  obtained  in  (3)  into  the 
units  used  in  measuring  the  lines  of  the  actual  figure. 

EXERCISES 

1.  A  man  walks  from  his  home  around  a  swamp  (Fig.  253). 
He  starts  from  his  home  at  A,  walks  0.95  mi.  north,  then  1.2  mi. 

east,  then  0.35  mi.  south.    How  far  from      B i2ml 

home  is  he  ? 

Solution.  Let  2  cm.  represent  1  mi.  Make  a 
drawing  on  squared  paper  of  the  distances  as 
shown  in  Fig.  253.  Then  on  the  squared  paper 
a  side  of  every  small  square  represents  0.1  mi. 
(Why  ?)  The  required  distance  is  the  number 
of  miles  represented  by  the  segment  A  D,  which 
is  13.9  small  units  long.  Hence  AD  represents  1.39  mi.  Why? 


348  GENERAL  MATHEMATICS 

2.  Show  how  the  four  steps  given  in  the  outline  of  Art.  395 
are  followed  in  the  solution  of  the  preceding  problem. 

3.  A  man  starting  at  a  point  S  walks  48  yd.  north  and  then 
56  yd.  east.    Find  the  direct  distance  from  the  stopping-point 
to  the  starting-point.    (Let  1  cm.  =  10  yd.) 

4.  A  man  walks  92  yd.  south,  then  154  yd.  east,  and  then 
132  yd.  north.    How  far  is  he  from  the  starting-point  ?    (Use 
1  cm.  for  every  12  yd.) 

5.  Two  men  start  from  the  same  point.    One  walks  15  mi. 
west,  then   9  mi.  north ;  the  other  walks  12  mi.  south,  then 
16  mi.  east.    How  far  apart  are  they  ? 

6.  Draw  to  scale  a  plan  of  your  desk  top  and  find  the 
distance  diagonally  across.    (Use  the  scale  1  cm.  =  1  ft.) 

7.  A  baseball  diamond  is  a  square  whose  side  is  90  ft. 
By  means  of  a  scale  drawing,  find  the  length  of  a  throw  from 
"  home  plate  "  to  "  second  base." 

8.  The    broken    line   ABC 
(Fig.  254)  represents    a   coun- 
try road.    Find  out  how  much 
nearer  it  would  be  to  walk  diag- 
onally across  country  from  A  to   A        3'3  mi>      B 

C  than  it  is  to  follow  the  road.  FlG-  254 

9.  A  roadbed  is  said  to  have  a  6%  grade  when  the  level 
of  the  road  rises  6  ft.  in  100  ft.  measured  horizontally. '  Draw 
to  scale  a  roadbed  520  yd.  long  which  has  a  6%  grade. 

10.  The  sides  of  a  triangular  chicken  lot  are  20  ft.,  16  ft., 
and  18  ft.  respectively.    Make  a  scale  drawing  of  this  lot  on 
squared  paper  and  estimate  the  area  by  counting  the  small 
squares  and  approximating  the  remaining  area. 

11.  In  a  map  drawn  to  the  scale  of  1  to  200,000  what  lengths 
will  represent  the  boundaries  of  a  rectangular-shaped  county 
40  mi.  long  and  20  mi.  wide  ?    Give  the  answer  to  the  nearest 
hundredth  of  an  inch. 


TRIGONOMETRY 


349 


12.  A  railroad  surveyor  wishes  to  measure  across  the  swamp 
at  A  B  represented  in  Fig.  255.  He  measures  the  distance  from 
a  tree  A  to  a  stone  C  and  finds  it  to  be  110  yd. 
Find  the  distance  across  the  swamp  if  the 
angle  at  C  is  70°  and  if  BC  =  100  yd.- 

NOTE.    The    lines    BC    and    AC    are    meas- 
ured by  means  of    a  steel   tape   (Fig.  256)   or  a 
surveyor's  chain  (Fig.  257).     The   angle  at  C  is 
measured  by  means  of  a  transit  (Fig.  43).    Chaining  pins  (Fig.  258) 
are  used  by  surveyors  to  mark  the  end-points  of  the  chain  or  tape. 


FIG.  256.    STEEL 
TAPE 


FIG.  257.   SURVEYOR'S 
CHAIN 


FIG.  258.    CHAIN- 
ING PINS 


*  13.  If  available  examine  a  steel  tape,  chain,  and  the  pins  used 

by  surveyors  and  report  to  class  on  the  length,  graduations,  etc. 

14.  In   Fig.  259,  S  represents  a  water-pumping  station  in 

Lake  Michigan.    A  and  B  represent  two  Chicago  buildings  on 


the  lake  shore.    Reproduce  the  measurements   to  scale  and 
determine  the  distance  of  S  from  each  of  the  two  buildings. 

15.  In  Fig.  260" a  swimming  course  AB  across  a  small  lake 
is  represented.    Find  A  B  by  means  of  a  scale  drawing. 


350  GENERAL  MATHEMATICS 

16.  A  triangular  lot  has  these  dimensions:  .4.6  =  20  yd.; 
BC  —  40  yd.;  A  C  =  30  yd.  Make  a  scale  drawing  of  the  lot  on 
squared  paper  and  determine  its  area.  (Since  the  formula  for 

the  area  of  a  triangle  A  =  —  calls  for  an  altitude,  the  student 

L 

will  draw  one  from  A  to  BC  and  then  apply  the  formula.) 

*17.  In  order  to  measure  the  distance  between  two  pump- 
ing stations  A  and  B  in  Lake  Michigan  a  base  line  C/>=18.8 
chains  long  (1  chain  =  66  ft.)  was  measured  along  the  shore. 
The  following  angles  were  then  measured:  Z.  A  CD  =  132° ; 
Z.BCD=  50°;  Z.CDA  =  46°;  ZCZ>£  =  125°.  Draw  the  figure 
to  scale  and  find  the  distance  from  A  to  B  in  feet. 

*18.  Two  streets  intersect  at  an  angle  of  80°.  The  corner 
lot  has  frontages  of  200  ft.  and  230  ft.  on  the  two  streets,  and 
the  remaining  two  boundary  lines  of  the  lot  are  perpendicular 
to  the  two  streets.  What  is  the  length  of  these  two  boundary 
lines  ?  What  is  the  area  of  the  lot  ? 

HINT.  Construct  the  two  perpendiculars  with  compasses.  Then 
draw  a  diagonal  so  as  to  form  two  triangles  and  construct  their 
two  altitudes  as  in  Ex- 16,  above. 

*19.  A  class  in  surveying  wishes 
to  determine  the  height  of  a  smoke- 
stack as  shown  in  Fig.  261.  The 
transit  is  placed  at  B,  and  the  angle 
ij  is  found  to  be  62°:  then  at  A, 
and  the  angle  x  is  found  to  l>e  32°. 

Line  AB  is  48  ft.  long  and  is  measured  along  level  ground. 
The  transit  rests  on  a  tripod  3^  ft.  high.  Find  the  height 
of  the  chimney. 

396.  Angle  of  elevation.  The  angles  x  and  y  which 
were  measured  in  Ex.  19  are  called  angles  of  elevation. 
The  angle  KAH  in  Fig.  262  shows  what  is  meant  by  an 
angle  of  .elevation.  To.  find  .the  angle  of  elevation,  the 


^x*  . 

A 

...-••'    /• 
/ 
B  ••-.  y 

I 

?B0n 

TRIGONOMETRY 

transit   is   placed   at   A   as   in    Fig.  262.    The   telescope 

of   the  transit   is   first   pointed    horizontally   toward    the 

smokestack.    The  farther  end  is  then  ^ 

raised  until  the  top  of  the  chimney 

K  is  in  the  line  of  sight.    The  angle 

KAH,    through    which    the    telescope    A  <0JLtal  Une  \{ 

turns,  is  the  angle  of  elevation  of   K 

.  .    ,  FIG.  262.    ANGLE  OK 

irom  A,  the  pouit  or  observation.  ELEVATION 


EXERCISES 

By  means    of   scale   drawings,  compasses,  and    protractor 
solve   the   following  exercises : 

1.  When  the  angle  of  elevation  of  the  sun  is  20°  a  building 
casts  a  shadow  82  ft.  long  on  level  ground.    Find  the  height  of 
the  building. 

2.  Find  the  angle  of  elevation  of  the  sun  when  a  church 
spire  80  ft.  high  casts  a  shadow  120  ft.  long. 

3.  A  roof  slopes  1  in.  per  horizontal  foot.    What  angle  does 
the  roof  make  with  the  horizontal  ? 

4.  A  light  on  a  certain  steamer  is  known  to  be  30  ft.  above 
the  water.    An  observer  on  the  shore  whose  instrument  is 
4  ft.  above   the  water  finds  the   angle   of  elevation  of  this 
light  to  be  6°.    What  is  the  distance  from  the  observer  to 
the  steamer  ? 

5.  What  angle  does  a  mountain  slope  make  with  a  horizontal 
plane  if  it  rises  150  ft.  in  a  horizontal  distance  of  one  tenth 
of  a  mile  ? 

6.  The  cable  of  a  captive  balloon  is  620  ft.  long.    Assuming 
the  cable  to  be  straight,  how  high  is  the  balloon  when^all 
the  cable  is  out  if,  owing  to  the  wind,  the  cable  makes  an  angle 
of  20°  with  the  level  ground  (that  is,  the  angle  of  elevation 
is  20°)  ? 


352  GENERAL  MATHEMATICS 

7.  On  the  top  of  a  building  is  a  flagpole.  At  a  point  A  on 
level  grour/4  70  ft.  from  the  building  the  angle  of  elevation  of 
the  top  of  the^lagpole  is  42°.  At  the  same  point,  A,  the  angle 
of  elevation  of  the  top  of  the  building  is  32°.  Find  the  height 
of  the  flagpole.  How  high  is  the  building  ? 

397.  Angle  of  depression.  A  telescope  at  M  in  the  top 
of  a  lighthouse  (Fig.  263)  is  pointed  horizontally  (zero 
reading),  and  then  the  farther  end  is  lowered  (depressed) 
until  the  telescope  points  to  a  boat  at  B.  The  angle  HMB, 
through  which  the  telescope  turns,  is  the  angle  of  depression 
of  the  boat  from  the  point  M.  In  Fig.  26  3,  Z  HMB  =  ^MBC. 
Why  is  this  true  ? 

EXERCISES 

1.  If   the    height  of  the   lighthouse  (Fig.  263)  is   220  ft. 
above  water,  and  the  angle  of  depression  of  the  boat,  as  seen 

from  M.  is  40°,  what  is  the  distance 

.„  .      .  H    horizontal  line    M 

of  the  boat  from  R  if  R  C  is  known 

to  be  40ft.? 

2.  A  boat  passes  a  tower  on  which 
is  a  searchlight  220  ft.  above  sea  level. 
Find    the    angle   through    which    the        B 

beam  of  light  must  be  depressed  from       FlG-  263-  ANGLE  OF 

,i     ,     •,  i  •  DEPRESSION 

the  horizontal  so   that  it  may  shine 

directly  on  the   boat  when    it   is   300  ft.   from   the    base  of 
the  tower. 

3.  How  far  is  the  boat  from  the  base  of  the  tower  if  the 
angle  of  depression"  is  51°  ?  30  °  ?    Xote  that  the  height  of  the 
lighthouse  is  known,  and  that  the  distance  of  a  boat  out  at  sea 
depends  on  the  size  of  the  angle ;  that  is,  the  distance  is  a 
function  of  the  angle.    In  other  words,  the  lighthouse  keeper 
needs  only  to  know  the  angle  of  depression  to  determine  the 
distance  of  a  boat  at  sea. 


TRIGONOMETRY 


353 


4.  From  the  top  of  a  mountain  2500  ft.  above  the  floor  of 
the  valley  the  angles  of  depression  of  two  barns  in  the  level 
valley  beneath,  both  of  which  were  due  east  of  the  observer, 
were  found  to  be  27°  and  56°.   What  is  the  horizontal  distance 
between  the  two  barns  ? 

5.  From  the  top  of  a  hill  the  angles  of  depression  of  two 
consecutive  milestones  on  a  straight  level  road,  running  due 
south  from  the  observer,  were  found  to  be  23°  and  47°.    How 
high  is  the  hill  ? 

398.  Reading  angles  in  the  horizontal  plane ;  bearing  of 
a  line.  The  special  terms  used  in  stating  problems  in  sur- 
veying and  navigation  make  the  problems  seem  unfamiliar, 
although  the  mathematics  is  frequently  not  more  difficult 


N 


N 


N 


N 


w- 


-EW- 


-E 


S  s  s  s 

FIG.  264.    ILLUSTRATING  THE  BEARING  OF  A  LINK 

than  iii  the  exercises  given  in  Art.  397.  Thus  the  acute 
angle  which  a  line  makes  with  the  north-south  line  is  the 
bearing  of  the  line. 

The  bearings  of  the  lines  indicated  by  the  arrows  in 
Fig.  264  are  read  as  follows :  75°  east  of  north,  46°  west 
of  north,  20°  east  of  south,  and  30°  west  of  south.  These 
are  written  more  briefly  as  follows:  N. 75° E.,  N.46°W., 
S.20°E.,  and  S.30°W. 


354 


GENEKAL  MATHEMATICS 


EXERCISES 

1.  Read  the  bearings  of  the  arrow  lines  in  Fig.  265. 

2.  WTith  a  ruler  and  protractor  draw  lines  having  the  fol- 
lowing bearings : 

(a)  26°  east  of  south.  (d)  37^°  west  of  south. 

(b)  39°  east  of  north.  (e)  33°  west  of  north. 

(c)  40°  west  of  north.  (f)  3°  east  of  south. 

3.  Write  in  abbreviated  form  the  bearings  of  the  lines  in  Ex.  2. 

N  N  N  N  ,V 

*f 


W- V — -E  W- —  — E  W-—- 


—-E  W—^- E  W- -h^-- 


s  s  s 

FIG.  265 

399.  Bearing  of  a  point.  The  bearing  of  a  point  B 
(Fig.  266)  from  a  point  0  is  the  bearing  of  the  line  OB 
with  reference  to  the  north-south  line  through  0. 

EXERCISES 

1.  In  Fig.  266  read  the  bear- 
ing of 

(a)  A  from  O.    (d)  0  from  B. 

(b)  0  from  A.    (e)  C  from  O. 

(c)  B  from  O.    (f)  0  from  C. 

2.  Point   A    is   6.4  mi.    east 
and  9.8  mi.  north  of  B.    Find 
the  distance  from  A  to  B.  What 
angle  does  AB  make  with  the 
north-south    line    through    B? 
What    is    the    bearing    of    B 
from  A?    of  A   from  B? 

3.  Sketch  the  figure  for  Ex.  2 

and  show  why  the  angles  appearing  as  results  for  Ex.  2  are  equal. 


TRIGONOMETRY  355 

4.  The  bearing  of  a  fort  B  from  A,  both  on  the  seacoast,  is 
N.  55°  W.  An  enemy's  vessel  at  anchor  off  the  coast  is  observed 
from  A  to  bear  northwest;  from  B,  northeast.  The  forts  are 
known  to  be  8  mi.  apart.  Find  the  distance  from  each  fort  to 
the  vessel. 

400.  The  limitations  of  scale  drawings.    By  this  time  the 
student  probably  appreciates  the  fact  that  a  scale  drawing 
has  its  limitations.     He  would  probably  not  agree  to  buy 
a  triangular  down-town  lot  whose  altitude  and  area  had 
been  determined  by  a  scale  drawing.    If  a  millimeter  on 
the  squared  paper  represents  0.1  of  a  mile,  a  slight  slip 
of  the  pencil  or  compasses  means  disaster  to  accuracy. 

Scale  drawing  is  used  extensively  by  the  surveyor  and 
engineer  in  the  following  ways:  (1)  as  a  method  of  esti- 
mating probable  results;  (2)  as  a  help  to  clear  thinking 
about  the  relations  of  lines  and  angles  involved  in  a  geomet- 
ric drawing  ;  (3)  as  a  valuable  check  on  results  obtained  by 
more  powerful  methods.  But  as  a  matter  of  fact  we  need  a 
more  refined  method  to  determine  lines  and  angles  where 
a  high  degree  of  accuracy  is  desirable.  We  shall  now  pro- 
ceed to  consider  a  far  more  efficient  method  of  determining 
such  lines  and  angles.  Most  students  will  find  the  method 
fascinating,  because  the  solution  is  simple  and  the  results 
obtained  are  as  accurate  as  the  lines  and  angles  which  are 
directly  measured. 

TRIGONOMETRY 

401.  Similar  right  triangles.    A  few  exercises  on  similar 
right  triangles  will  help  the  student  to  understand  the 
new  and  more  accurate  method  of  determining  lines  and 
angles.    This  method  may  be  used  independent  of  scale 
drawings,  is  shorter  in  most  cases,  and  lays  the  foundation 
for  future  mathematical  work. 


GENERAL  MATHEMATICS 


a 


b          O 

<;.  207 


EXERCISES 

1.  With    the    protractor   construct  a  right-angled  triangle 
having  an  angle  of  37°.    Letter  the  figure  as  suggested   in 
Fig.  267.    Measure  the  lines  a,  b,  and  c.    Let 

2  cm.  represent  1  unit.    Find  the  value  of  the 

ratios  - »  -  >  and  —  to  two  decimal  places. 
c     c  l> 

2.  Compare  your  result  with  other  mem- 
bers of  the  class.   Did  all  members  of  the  class 
use  the  same  length  for  the  bases  ?  Are  any 

two  of  the  triangles  drawn  necessarily  of  the  same  size  ?   Show 

why  the  result  obtained  for  — 

c 

should  be  the  same  number  as 
the  results  of  your  classmates. 

3.  Prove  that  if  two  right 
triangles  have  an  acute  angle 

of  one  equal  to  an  acute  angle  of  the  other,  the  ratios  of  their  corre- 
sponding sides  are  equal.  Write  two  proportions.  (Use  Fig.  268.) 

4.  Could  you  draw  a  right  triangle  with  angle  A  =37°  in 

which  -  does  not  equal  approximately  0.60,  or  ~  ?    Prove. 

HINT.    The  fact  that  -  =  -  means  that  in  every  right  triangle  the 
side  opposite  a  37°  angle  is  approximately  £  as  long  as  the  hypotenuse. 

5.  A  balloon  B  (Fig.  269)  is  fastened  by  a  cable  200  ft.  long. 
Owing  to  the  wind  the  cable  is  held  practically  straight  and 
makes  a  37°  angle  with  the  horizontal.  How  high  is  the  balloon  ? 

Solution.    This  triangle  is  similar  to  every  tri-  B 

angle  drawn  by  the  class  in  Ex.  1.    Prove. 


Therefore 


—  =  O.GO. 
200 

a  =  120  ft. 


\37 


90; 


Solving, 

Note  that  the  solution  is  exceedingly  simple 
(only  two  equations)  and  that  the  accuracy  of 
the  result  does  not  now  depend  upon  the  accuracy  of  Fig.  269. 


b 
FIG.  269 


TRIGONOMETRY  357 

402.  Sine  of  an  angle.   The  ratio  -  (Fig.  270)  is  called  the 
sine  of  the  angle  A.  The  abbreviation  for  "sine"  is"sin."  This 

definition  may  be  written  sin  A  —  -  •    Thus, 

c 

sin  37°  =  §  =  O.GO  (approx.).    Do  you  think ' 

we  would  have  obtained  the  same  value  for  -       

c     A        b      C 

if  in  Ex.  1  we  had  made  the  angle  47°  ?         FHJ.  270 

EXERCISES 

1.  Find  the  sine  of  20°,  using  the  definition  given  in  Art.  402. 
HINT.    As  in  Ex.  1,  Art.  401,  construct  the  triangle,  measure  a 

and  c,  and  find  the  value  of  -  to  two  decimal  places. 

c 

2.  Find  the  sine  of  each  of  the  following  angles :  10°,  15°, 
2u°,  32°,  47°,  68°,  87°.    Compare  each  result  with  the  results 
of  your  classmates. 

403.  Table  of  sines.    The  preceding  exercises  show  that 
the   sine   of   the   angle  changes  with  the   angle ;   that  is, 
sin  68°  is  not  equal  to  sin  37°.    By  taking  a  large  sheet  of 
graphic  paper  and  a  very  large  unit  we  could  get  a  fairly 
good  table,  but  it  would  be  too  much  trouble  to  do  'this 
for  every  problem.    Such  a  table  has  been  very  carefully 
calculated  for  you   in  the   first  column   of  the  table    in 
Art.  410. 

EXERCISE 

Turn  to  the  table  in  Art.  410  and  .see  how  efficient  yo\i 
have  been  by  comparing  your  results  for  Ex.  2,  Art.  402,  with 
the  table. 

404.  Cosine  of  an  angle.    The  exercises  given   in   this 
article  will  introduce  another  trigonometric  ratio. 


358  GEJSEKAh  MATHEMATICS 

INTRODUCTORY  EXERCISES 

1.  Construct    a    right-angled   triangle    with   angle    .1    (see 
Fig.  270)  equal  to  43°.    Measure  b  and  c.    Find  the  quotient 

-.to  two  decimal  places.  Compare  the  results  with  those  of 
the  other  members  of  the  class. 

2.  Show  that  all  results   ought  to  agree  to  two  decimal 
places.   The  ratio  -  (Fig.  270)  is  called  the  cosine  of  the  angle  A . 

C 

The  abbreviation  for  "  cosine  "  is  "  cos."  Thus,  cos  43°  =  0.73 
(approx.).  This  means  that  in  an}-  right-angled  triangle  the 
side  adjacent  to  the  angle  43°  is  about  -j7^-  as  long  as  the 
hypotenuse. 

3.  Find  the  cosine  of  5°,  18°,  25°.  35°,  47°,  65°,  87°. 

4.  Compare  the  results  for  Ex.  3  with  the  table  of  cosines 
in  Art.  410. 

405.  Tangent  of  an  angle.  We  shall  now  introduce  a 
third  important  ratio  connected  with  similar  right  triangles. 
Historically  the  tangent  ratio  came  first.  We  shall  have 
occasion  to  learn  more  about  it. 

INTRODUCTORY  EXERCISES 

1.  In  Fig.  270,  what  is  the  value  of  -  ?  Compare  your  result 
with  the  results  obtained  by  other  members  of  the  class. 

2.  Show  that  all  the  results  obtained  for  -  in  Ex.  1  should 

b 
agree. 

The  ratio  -  is  called  the  tangent  of  angle.  A.  In  speak- 
ing of  the  tangent  of  43°  we  mean  that  the  side  opposite 
angle  A  is  y9^  (approx.)  of  the  length  of  the  side  adjacent. 
The  abbreviation  for  "  tangent"  is  "tan."  Thus,  tan  45°  =  1. 


TRIGONOMETRY  359 

EXERCISES 

1.  Find  the  tangent  of  11°,  36°,  45°,  57°,  82°. 

2.  Compare  the  results  of  Ex.  1  with  the  table  of  tangents 
in  Art.  410. 

406.  Trigonometric  ratios.   Solving  a  triangle.  The  ratios 

a    b         ,  a 

->  ->  and  -  are  called  trigonometric  ratios.    We  shall  now 

proceed    to    show   that   the    use    of    these    ratios   greatly 

simplifies  the  solution  of  many  problems  in- 

volving indirect  measurements.    By  their  use 

any  part  of  a  right  triangle  can  be  found  if 

any  two  parts  (not  both  angles)  besides  the 

right  angle  are  given.    This  process  is  called 


solving  the  triangle. 

FIG.  271 

407.  Summary  of  definitions.   The  follow- 

ing  outline  will  be  found  convenient  to  the  student  in 
helping  him   to  remember  the  definitions  (see  Fig.  271): 

a       side  opposite 
1.   sm  A=-  —  —  -  £C- 

c         lii/potennse 

b      side  adjacent 

'2.  cos  A  =  -  =  —  —  —  • 

i'         hypotenuse 

a       side  opposite 

3.  tan  A  =  •-  =  - 

o      side  adjacent 

408.  Trigonometric  ratios  clear  examples  of  the  function 
idea.    Either  by  your  own  crude  efforts  at  building  a  table 
or  by  a  study  of  the  table  of  ratios  given,  it  is  easy  to  see 
that  the  value  of  the  ratio  changes  as  the  angle  changes: 
that  is,  a  trigonometric  ratio  depends  for  its  value  upon 


360  GENERAL  MATHEMATICS 

the  size  of  the  angle.  Hence  the  ratios  furnish  us  with 
one  more  clear  example  of  the  function  idea.  We  may 
therefore  refer  to  them  as  trigonometric  functions. 

HISTORICAL  NOTE.  Trigonometric  ratios  are  suggested  even 
in  the  Ahmes  Papyrus  (c.  1700  B.C.  ?),  which,  as  has  been  stated, 
may  itself  be  a  copy  of  some  other  collection  written  before  the 
time  of  Moses.  In  dealing  with  pyramids  Ahmes  makes  use  of 
one  ratio  that  may  possibly  correspond  roughly  to  our  cosine  and 
tangent. 

The  first  to  make  any  noteworthy  progress  in  the  development  of 
trigonometry  was  Hipparchus,  a  Greek,  who  lived  about  150  n.r. 
He  studied  at  Alexandria,  and  later  retired  to  the  island  of  Rhodes, 
where  he  did  his  principal  work.  He  was  able  to  calculate  the  length 
of  a  year  to  within  six  minutes. 

The  Hindus  contributed  to  the  early  development  of  the  science, 
from  about  A.D.  500,  and  the  Arabs  added  materially  to  their  work 
from  about  A.D.  800  to  A.D.  1000 

Regiomontanus  (or  Johann  Miiller,  1436-1476),  a  German,  freed 
the  subject  from  its  direct  astronomical  connection  and  made  it  an 
independent  science. 

In  the  sixteenth  century  the  subject  developed  slowly,  but  in  the 
seventeenth  century  it  made  a  very  decided  advance,  due  to  the 
invention  of  logarithms,  mentioned  later,  and  to  the  great  improve- 
ment of  algebraic  symbolism  which  made  it  possible  to  write  trigo- 
nometric formulas  in  a  simple  manner.  Trigonometry  in  the  form 
that  we  know  it  may  be  said  to  have  been  fully  developed,  except 
for  slight  changes  in  symbols,  in  the  seventeenth  century. 

409.  Table  of  trigonometric  ratios  of  angles  from  1°  to  89°. 
The  student  should  now  become  familiar  with  the  table 
on  the  following  page.    The  ratios  are  in  most  cases  only 
approximate,   but   are    accurate    enough  for   all   ordinary 
work. 

410.  The  use  of  a  trigonometry  table.    The  problems 
beginning  on  page  362  are  intended  to  furnish  the  student 
practice  in  the  use  of  the  table. 


TRIGONOMETRY 


361 


Angle 

Sine 

(oPP.\ 
\hvp.) 

Cosine 
tad±\ 
\hupj 

Tangent 
(opp.\ 
\adj.) 

Angle 

Sine 
(oM>.\ 
\hvP.J 

Cosine 
/adJA 
\hyp.) 

Tangent 

(opp.\ 
\adj.) 

0° 

1° 
2° 
3° 
4° 

.000 
.017 
.035 
.052 
.070 

1.000 
1.000 
.999 
.999 
.998 

.000 
.017 
.035 
.052 
.070 

45° 
46° 
47° 
48° 
49° 

.707 
.719 
.731 
.743 

.755 

.707 
.695 
.682 
.669 
.656 

1.000 
1.036 
1.072 
1.111 
1.150 

5° 
6° 
7° 
8° 
9° 

.087 
.105 
.122 
.139 
.156 

.996 
.995 
.993 
.990 
.988 

.087 
.105 
.123 
.141 

.158 

50° 
51° 
52° 
53° 
54° 

.766 
.777 
.788 
.799 
.809 

.643 
.629 
.616 
.602 
.588 

1.192 
1.235 
1.280 
1.327 
1.376 

10° 
11° 
12° 
13° 
14° 

.174 
.191 
.208 
.225 
.242 

.985 
.982 
.978 
.974 
.970 

.176 
.194 
.213 
.231 
.249 

55° 
56° 
57° 
58° 
59° 

.819 

.829 
.839 
.848 
.857 

.574 
.559 
.545 
.530 
.515 

1.428 
1.483 
1.540 
1.600 
1.664 

15° 
16° 
17° 
18° 
19° 

.259 
.276 
.292 
.309 
.326 

.966 
.961 
.956 
.951- 
.946 

.268 
.287 
.306 
.325 
.344 

60° 
61° 
62° 
63° 
64° 

.866 
.875 
.883 
.891 
.899 

.500 
.485 
.469 

.454 
.438 

1.732 
1.804 
1.881 
1.963 
2.050 

20° 
21° 
22° 
23° 
24° 

.342 
.358 
.375 
.391 

.407 

.940 
.934 
.927 
.921 
.914 

.364 
.384 
.404 
.424 
.445 

65° 
66° 
67° 
68° 
69° 

.906 
.914 
.921 
.927 
.934 

.423 
.407 
.391 
.375 

.358 

2.145 
2.246 
2.356 
2.475 
2.605 

25° 
26° 
27° 
28° 
29° 

.423 
.438 
.454 
.469 
.485 

.906 
.899 
.891 
.883 

.875 

.466 
.488 
.510 
.532 
.554 

70° 
71° 
72° 
73° 

74° 

.940 
.946 
.951 
.956 
.961 

.342 
.326 
.309 
.292 
.276 

2.747 
2.904 
3.078 
3.271 
3.487 

30° 
31° 
32° 
33° 
34° 

.500 
.515 
.530 
.545 
.559 

.866 
.857 
.848 
.839 
.829 

.577 
.601 
.625 
.649 
.675 

75° 
76° 
77° 
78° 
79° 

.966 
.970 
.974 
.978 
.982 

.259 
.242 
.225 
.208 
.191 

3.732 
4.011 
4.331 
4.705 
5.145    . 

35° 
36° 
37° 
38° 
39° 

.574 
.588 
.602 
.616 
.629 

.819 
.809 
.799 

.788 
.777 

.700 
.727 
.754 
.781 
.810 

80° 
8.1° 
82° 
83° 
84° 

.985 
.988 
.990 
.993 
.995 

.174 
.156 
.139 
.122 
.105 

5.671 
6.314 
7.115 
8.144 
9.514 

40° 
41° 
42° 
43° 
44° 

.643 
.656 
.669 
.682 
.695 

.766 
.755 
.743 
.731 
.719 

.839 
.869 
.900 
.933 
.966 

85° 
86° 
87° 
88° 
89° 

.996 
.998 
.999 
.999 
1.000 

.087 
.070 
.052 
.035 
.017 

11.430 
14.301 
19.081 
28.636 
57.290 

45° 

.707 

.707 

1.000 

90° 

1.000 

.000 

NOTE.  The  abbreviation  hyp.  means  "hypotenuse";  adj.  means  "the 
side  adjacent  to  the  angle";  opp.  means  "the  side  opposite  the  angle." 


362  GENEliAL  MATHEMATICS 

EXERCISES 

1.  A  balloon  B  (Fig.  272)  is  anchored  to  the  ground  at  a 
point  .1  by  a  rope,  making  an  angle  of  57°  with  the  ground. 
The  point  C  on  the  ground  directly  under  the  balloon  is  146  ft. 
from  A.    Assuming  the  rope  to  be  straight,  find 
the  height  of  the  balloon. 

Solution.    Let  a  =  height  of  balloon. 

Then  -—  =  tangent  of  57°. 

146 

But  by  the  table,  Art.  410, 

tan  57°  =  1.54. 


Hence  -£-  =  1.51.  A         146/      ° 

Fir    272 
Solving  for  a.  a  =  224.84  ft. 

NOTK.  The  figure  does  not  need  to  be  drawn  accurately,  for  our 
results  are  obtained  independently  of  it.  The  solution  is  brief  and 
depends  for  its  accuracy  upon  the  accuracy  of  the  angle  57°,  the  accu- 
racy of  the  length  of  the  line  A  C,  and  the  accuracy  of  the  tangent  table. 

2.  The  angle  of  elevation  of  an  aeroplane  at  a  point  A  on 
level   ground   is   53°.     The   point  C  on   the  ground  directly 
under  the  aeroplane  is  315  yd.  from  A.   Find  the  height  of 
the  aeroplane. 

3.  The  length  of  a  kite  string  is  210yd.  and  the  angle  of 
elevation   of  the   kite  is   48°.     Find  the  height  of  the  kite, 
supposing  the  line  of  the  kite  string  to  be  straight. 

4.  A  pole  20  ft.  in  length  stands  vertically  in  a  horizontal 
area,  and  the  length  of  its  shadow  is  16.78  ft.    Find  the  angle 
of  elevation  of  the  sun. 

HINT.    Find  the  value  of  the  tangent  -•    Then  look  in  the  table 

o 

to  see  what  angle  has  a  tangent  corresponding  to  the  value  of 

-•   It  may  be  necessary  for  you  to  approximate,  since  the  table  is 

b 

not  calculated  for  minutes.    Ask  your  instructor  to  show  you  a  more 

complete  table  of  trigonometric  ratios. 


TRIGONOMETRY 


5.  A  tree  is  broken  by  the  wind  so  that  its  two  parts  form 
with  the  ground  a  right-angled  triangle.    The  upper  part  makes 
an  angle  of  55°  with  the  ground,  and  the  distance  on  the 
ground  from  the  trunk  to  the  to.p  of  the  tree  is  57  ft.    Find 
the  length  of  the  tree. 

6.  A  circular  pool  has  a  pole  standing  vertically  at  its 
center,  and  its  top  is  50  ft.  above  the  surface.    At  a  point  in 
the  edge  of  the  pool  the  angle  subtended  by  the  pole  is  25°. 
Find  the  radius  and  the  area  of  the  pool. 

7.  A  ladder  35  ft.  long  leans  against  a  house  and  reaches 
to  a  point  19.6  ft.  from  the  ground.    Find  the  angle  between 
the  ladder  and  the  house  and  the  distance  the  foot  of  the 
ladder  is  from  the  house. 

8.  Measure  two  adjacent  edges  of  your  desk  or  of  a  rec- 
tangular table,  say  your  study  table.   Find  the  angles  that  the 
diagonal  makes  with  the  edges  (1)  by  drawing  an  accurate 
figure   and   measuring  the   angle  with  a   pro- 
tractor ;  (2)  by  use  of  the  trigonometric  ratios. 

9.  The  tread  of  a  step  on  a  certain  stairway 
is  11  in.  wide  ;  the  step  rises  8  in.  above  the  next 
lower  step.  Find  the  angle  at  which  the  stairway 
rises  (1)  by  means  of  a  protractor  and  an  accurate 
figure  ;  (2)  by  means  of  a  trigonometric  ratio. 

10.  To  find  the  distance  across  a  lake  (Fig.  273) 

between  two  points  A  and  C,  a  surveyor  measured  off  71  ft.  on 
a  line  EC  perpendicular  to  A  C.  He  ^,Z> 

then  found  ZC^£=53°.  Find  A  C. 

11.  The  Washington  Monument 

is  555  ft.  high.    How  far  apart  are   g         x        A  C 

two  observers  who  from  points  due  Fjo  274 

west  of  the  monument  observe  its 

angles  of  elevation  to  be  20°  and  38°  respectively  ?  (See  Fig.  274.) 
HINT.  Find. If. 


FIG.  273 


Then 


x  +  value  of  A  C 


=  tan  20° 


364 


GENERAL  MATHEMATICS 


*12.  A  man  standing  on  the  bank  of  a  river  observes  that  the 
angle  of  elevation  of  the  top  of  a  tree  on  the  opposite  bank  is 
56° ;  when  he  retires  55  ft.  from  the  edge  of  the  river  the 
angle  of  elevation  is  32°.  Find  the  height  of  the  tree  and  the 
width  of  the  river. 

*13.  From  the 
summit  of  a  hill 
(Fig.  275)  there  are 
observed  two  con- 
secutive milestones  on  a  straight  horizontal  road  running  from 
the  base  of  the  hill.  The  angles  of  depression  are  found  to  be 
13°  and  8°  respectively.  Find  the  height  of  the  hill. 


FIG.  275 


HINT.    Construct  TC  _L  CMZ. 
Let  CMl  =  x. 


Then 


-  =  tan  77°,     (Why?) 
h 


and 


x  +  1 
h 


=  tan  82°.     (Why?) 


(1) 
(2) 


Subtracting  (1)  from  (2),       -  =  tan  82°  -  tan  77°. 
Consult  the  table  on  page  36,  substitute,  and  solve  for  /;. 

*14.  A  railroad  having  a  hundred-foot  right  of  way  cuts 
through  a  farmer's  field  as  shown  in  Fig.  276.  If  the  field  is 
rectangular  and  the  measurements 
are  made  as  shown,  find  the  num- 
ber of  square  rods  occupied  by  the 
right  of  way  and  the  assessed  damage 
if  the  land  is  appraised  at  $200 

an  acre. 

FIG.  276 

.15.  A  ship  has  sailed  due  south- 
west a  distance  of  2.05  mi.    How  far  is  the  ship  south  of 
the  starting  point  ?    How  far  is  it  west  of  the  starting  point  ? 


16.  From  the  top  of  a  mountain  4260  ft.  above  sea  level  the 
angle  of  depression  of  a  distant  boat  is  41°.  How  far  is  the 
boat  from  the  summit  of  the  mountain  ? 

*17.  Sketch  the  figure  and  solve  the  right-angled  triangle 
ABC  when 

(e)    I  =  92.5°,  c  =  100  ft. 
(f  )   a  =  15.2°,  c  =  50  ft. 
(g)  .1  =  40°,     c  =  80  ft. 
(h)  B  =  82°,     c  =  100  ft. 


(a)  A  =  30°,  a  =  30  ft. 
.  (b)  B  =  42°,  b  =  60  ft. 

(c)  A  =  64°,  f,  =  22  ft. 

(d)  a  =  35°,  I  =  85  ft. 


411.  A  trigonometric  formula  for  the  area  of  a  triangle. 

It  can  be  shown  that  the  area  of  a  triangle  equals  half  the 

product  of  any  two  sides  multiplied  by 

the  nine  of  the  included  angle  ;  that  is, 

ab  sin  A 

~2~ 

Solution.    In   Fig.  277  construct  the 
altitude  CD. 

Then  T  (the  area)  =  y.      (Why?)  (1) 

But  -  =  sin  A  (see  the  definition  of  "sine").  (2) 


Whence  h=bsinA.     (Why?) 

Substituting  the  value  of  7t  in  (1), 

_  be  sin  A 
i  —  — ^       * 


(3) 


EXERCISES 

1.  A  boy  discovers  that  his  father's  drug  store  completely 
covers  their  triangular  lot  and  that  it  extends  60  ft.  and  80  ft. 
on  two  sides  from  a  corner.  With  a  field  protractor  he  measures 
the  angle  between  the  streets  and  finds  it  to  be  58°.  He  then 
tries  to  find  the  area  of  the  lot.  What  result  should  he  get  ? 
*  2.  Prove  that  the  area  of  a  parallelogram  equals  the  product 
of  two  sides  and  the  sine  of  the  angle  included  between  these 
two  sides. 


366  GENERAL  MATHEMATICS 

SUMMARY 

412.  This  chapter  has  taught  the  meaning  of  the  fol- 
lowing words  and  phrases :  scale  drawing,  surveyor's  chain, 
steel  tape,  angle  of  elevation,  angle  of  depression,  bearing 
of  a  line,  bearing  of  a  point,  sine  of  an  angle,  cosine  of  an 
angle,  tangent  of  an  angle,  trigonometric  ratios  or  trigo- 
nometric functions,  solving  a  triangle. 

413.  Scale  drawings  were  used  as  a  means  of  indirect 
measurement. 

414.  A    scale    drawing    is    useful   in   making  estimates 
of  angles,  lines,  and  areas,  in  getting  a  clear  picture  in 
mind  of  the  relation  of  the  parts  that  make  up  the  figure, 
and  in   checking   the   accuracy   of  an   algebraic   solution. 
However,  it  is  not  as  brief  and  accurate  as  the  algebraic- 
sol  ution. 

415.  If  two  right  triangles  have  an  acute  angle  of  one 
equal  to  an  acute  angle  of  the  other,  the  ratios  of  their 
corresponding  sides  are  equal. 

416.  The  chapter  contains  a  table  of  trigonometric  ratios 
of  angles  from  1°  to  89°  and  .correct  to  three  decimal  places. 

417.  Trigonometric  ratios  furnish  us  with  a  powerful 
method  of  solving  triangles. 

418.  The  area  of  a  triangle  may  be  expressed  by  the 

,.          ,             be  sin  A 
formula  T=  — — 

419.  The  area  of  a  parallelogram  equals  the  product  of 
two  sides  and  the  sine  of  the  angle  included  between  these 
two  sides. 


CHAPTER  XV 


THEORY  AND  APPLICATION   OF   SIMULTANEOUS  LINEAR 
EQUATIONS ;  CLASSIFIED  LISTS  OF  VERBAL  PROBLEMS 

420.  Two  unknowns ;  solution  by  the  graphic  method. 
In  solving  verbal  problems  it  is  sometimes  desirable  to 
use  two  unknowns.  This  chapter  aims  to  teach  three 
methods  which  the  pupil  may  apply  to  such  problems.  The 
graphic  method  is 
shown  in  the  dis- 
cussion of  the  fol- 
lowing problem: 

In  a  baseball  game 
between  the  Chicago 
Cubs  and  the  New 
York  ,  Giants,  the 
Cubs  made  four 
more  hits  than  the 
Giants.  How  many 
hits  did  each  team 
make? 


-1-5- 


-10 


•x- 


Fir..  278 


If  we  let  x  repre- 
sent   the    number   of 
hits  made  by  the  Cubs 
and  y  the  number  made  by  the  Giants,  then  the  equation  x  =  y  +  4 
expresses  the  condition  as  set  forth  in  the  problem. 

Obviously  there  are  any  number  of  possible  combinations  such 
that  the  number  of  hits  made  by  one  team  may  be  four  more  than 
the  number  made  by  the  other  team.  This  is  clearly  shown  in  the 
graph  of  the  eqxiation  x  =  ?/  +  4  in  Fig.  278. 

367* 


3b'S 


GENERAL  MATHEMATICS 


EXERCISES 

1.  From  the  graph   in  Fig.  278  find  the  number  of  hits 
made  by  the  Giants,  assuming   that  the  Cubs   made  6;   8; 
10;  15;  20. 

2.  Show  that  every  point  (with  integral  coordinates)  on  the 
line  will  give  a  possible  combination  of  hits  such  that  x  —  y  +  4. 

NOTE.    By  this  time  the   student  is  no  doubt  convinced  that  a 
definite  solution  of  the  problem  as  stated  is  impossible,  because  it 
involves  two  unknowns  and  we  have  given  but  one  fact.   Another  fact 
which    should    have    been 
included    in    the    problem 
is  that  the  total  number  of 
hits  made  by  both  teams 
was  18. 

If  we  write  the  equa- 
tion x  +  y  =18,  expressing 
'this  second  fact,  and  study 
it  by  means  of  the  graph, 
Fig.  279,  we  see  that  there 
is  more  than  one  possi- 
ble combination  such  that 
the  total  number  of  hits 
made  is  18. 


-20 


-10 


FIG. 279 


3.  Find  from  the  graph 
in  Fig.  279  the  number 

of  hits  made  by  the  Giants  if  the  Cubs  made  4 ;  6 ;  9 ;  12 ;  15. 

4.  Show  .that  every  point  (with  integral  coordinates)  on  the 
line  will  give  a  possible  solution  for  the  equation  x  +  y  =  18. 

XOTE.  We  have  not  been  able  to  obtain  a  definite  solution, 
because  we  have  been  considering  the  two  facts  about  the  ball  game 
separately.  The  two  equations  express  different  relations  between 
the  two  unknowns  in  the  ball  game.  This  means  that  we  must  find 
one  pair  of  numbers  which  will  satisfy  both  equations,  and  to  do 
this  we  must  graph  both  equations  on  the  same  sheet  to  the  same 
scale,  as  shown  in  Fig.  280. 


SIMULTANEOUS  LINEAR  EQUATIONS        369 


5.  Find  a  point  in  the  graph  of  Fig.  280  that  lies  on  both 
lines.    What  does  this  mean  ? 

6.  What  are  the  a:- values  and  y- values  of  this  point  ? 

7.  Show  that  the  pair  of  values  obtained  in  Ex.  6  will  satisfy 
both  equations. 


(1) 


£ 

y 

4 

0 

9 

5 

20 

16 

-20 


x  =  y  +  4,  (1) 
x  +  y  =  18.  (2) 


FIG.  280.   THE  GRAPH  OF  A  PAIR  OF  SIMUL- 
TANEOUS LINEAR  EQUATIONS 


The  preceding  exercises  show  that  the  point  of  inter- 
section, namely  x  =  11,  y  =  7,  is  on  both  lines,  and  hence 
the  solution  of  the  problem  is  x  =  11,  y  =  7.  Thus  the 
number  of  hits  made  by  the  Cubs  was  11,  and  the  number 
made  by  the  Giants  was  7. 

421.  Simultaneous  linear  equations.  A  pair  of  linear  equa- 
tions which  are  satisfied  at  the  same  time  (simultaneously) 
by  the  same  pair  of  values  are  called  simultaneous  linear 
equations.   The  graph  is  a  pair  of  intersecting  straight  lines. 

422.  System   of   equations.    A    pair   of   equations    like 
those  in  Art.  420   is   often   called  a  system  of  two  linear 
equations.    A  system  of  linear  equations  having  a  common 


'61  it  GENERAL   MATHEMATICS 

solution  is  said  to  be  solved  when  the  correct  values  of  the 
unknowns  are  determined.  In  the  graphic  method  the 
coordinates  of  the  point  of  intersection  furnish  the  solu- 
tion. The  following  is  a  summary  of  the  graphic  method 
of  solving  a  pair  of  simultaneous  linear  equations: 

1.    Graph  loth  equations  to  the  same  scale. 
'2.    Find  the  point  of  intersection  of  the  two  Hnrs  obtained 
in  1. 

3.  Estimate  as  accurately  as  possible  the  x-raluc  find  the 
y-value  of  this  point. 

4.  Check  by  substituting  in  both  equation*. 

EXERCISES 

1.   Solve  the  following  systems  by  the  graphic  method  and 
check  each  : 

M*  +  y=s7'  ,  v2x  +  3y  =  23,  5y-3*  =  19, 

W-  ')  } 


3*  +  2^=27,  2y  +  3x=13, 

^5z-4*/  =  l.        (    ;  5y-6x=-S.         ' 

2.  What  difficulties  did  you  have  in  finding  the  correct 
values  for  the  coordinates  of  the  points  of  intersection  in  the 
problems  of  Ex.  1  ? 

423.  Indeterminate  equations.  A  single  equation  in  two 
unknowns  is  satisfied  by  an  infinite  (unlimited)  number 
of  values,  but  there  is  no  one  pair  of  values  which  satisfy 
it  to  the  exclusion  of  all  the  others  ;  for  example,  the  equa- 
tion x  +  y  =  4  is  satisfied  by  as  many  pairs  of  values  as 
are  represented  by  each  distinct  point  on  the  graph  of 
the  equation  x  +  y  =  4.  Such  an  equation  is  called  an 
in  determinate  equation. 


SIMULTANEOUS  LINEAR  EQUATIONS        371 


EXERCISES 

1.  Find  three  solutions  for  each  of  the  following  indeter- 
minate equations : 

(a)  x  +  y  =  7.         (c)  y  —  z  =  6.          (e)  5  x  —  z  =  2. 
(b)m  +  3n  =  5.    (d)  2 x  -  4  ?/  =  3.    (f)  3z  -  4  >/ -1=  0. 

424.  Contradictory  equations.  It  sometimes  happens  that 
even  though  we  have  two  equations  in  tyvo  unknowns,  it 
is  still  impossible  to  obtain  a  distinct  or  a  unique  solution, 
as  is  shown  by  the  following  example: 

Find  two  numbers  such  that  their  difference  is  12  arid  such 
that  twice  the  first  diminished  by  twice  the  second  is  equal  to  14. 


-20 


-1-0 


20- 


&Z 


f 


FIG.  281.  THE  GRAPH  OF  A  PAIR  OF  CONTRADICTORY  EQUATIONS 

If  we  let  x  denote  one  number  and  y  the  other,  then  from  the 
first  condition, 


x  -  y  =  12. 
From  the  second  condition,  '2  .c  —  '2  y  —  14. 


(1) 

(2) 


In  order  to  study  the  problem  fxirther  we  will  construct  the 
graphs  of  (1)  and  (2)  with  reference  to  the  same  coordinate 
axes  (Fig.  281). 


372  GENERAL  MATHEMATICS 

EXERCISES 

1.  What  relation  seems  to  exist  between  the  two  lines  of 
the  graph  in  Fig.  281  ? 

2.  Are  there,  then,  any  two  numbers  which  will  satisfy  the 
conditions  of  the  problem  given  on  the  preceding  page  ? 

A  system  of  equations  which  expresses  a  contradictory 
relation  between  the  unknowns  is  called  a  system  of 
contradictory,  or  inconsistent,  equations.  The  graph  consists 
of  two  (at  least)  parallel  lines.  The  definition  suggests 
that  in  a  verbal  problem  one  of  the  given  conditions  is 
not  true. 

425.  Identical  equations.  A  type  of  problem  which  has  no 
unique  solution  but  admits  of  many  solutions  is  illustrated 
by  the  following  problem : 

Divide  a  pole  10  ft.  long  into  two  parts  so  that  3  times  the 
first  part  increased  by  3  times  the  second  part  is  equal  to  30. 

If  we  let  x  and  ?/  denote  the  length  of  the  two  parts,  the  conditions 
of  the  problem  are  represented  by  the  equations 

x  +  y  =  10,  (l) 

and  3  x  +  3  y  =  30.  (2) 

EXERCISES 

1.  Graph  the  equations  x  +  y  =  10  and  3  x  +  3  y  =  30  to 
the  same  scale.    Interpret  the  graph. 

2.  Divide  the  equation  3sc  +  3?/  =  30  by  3  and  compare 
the  result  with  the  equation  x  -\-  y  =  10. 

Equations  like  (1)  and  (2),  above,  which  express  the 
same  relation  between  the  unknowns,  are  called  identical, 
dependent,  or  equivalent  equations.  Like  an  indeterminate 
equation,  they  have  an  infinite  number  of  solutions  but 


SIMULTANEOUS  LINEAR  EQUATIONS        373 

no  distinct  solution.  Their  graphs  coincide.  If  a  verbal 
problem  leads  to  two  identical  equations,  one  condition 
has  been  expressed  in  two  different  ways. 

426.  Outline  of  systems  of  equations  and  their  number  of 
solutions.  We  have  seen  that  a  linear  system  of  equations 
in  two  unknowns  may  be 

1.  Determinant  and  have  a  distinct  solution.    (Tlie  lines 
intersect.) 

2.  Contradictory  and  have  no  distinct  solution.    {The  lines 
are  parallel.} 

3.  Identical  and  have  an  infinite  number  of  solutions.    (  The 
lines  are  coincident.} 

EXERCISES 

1.  Classify  the  following  systems  according  to  the  preceding 
outline  by  drawing  graphs  of  each  system  : 


=    ,  -     =    , 

;  6  x  +  8  y  =  10.  ;  2  x  +  3  y  =  3. 

2.  Could  you  have  classified  the    four  systems  in  Ex.  1 
without  graphing  them  ?    Explain. 

427.  Algebraic  methods  of  solving  systems  of  equations  in 
two  unknowns.   It  is  often  difficult  (sometimes  impossible) 
to  judge  the  exact  values  in  a  graphic  solution.   The  graphic 
method  helps  us  to  see  what  is  meant  by  a  solution,  but 
i,t  is  not,  in  general,  as  exact  and  concise  a  method  as  the 
algebraic  methods  which  we  shall  now  illustrate. 

428.  Elimination.    To  solve  a  system  without  the  use 
of  graphs  it  will  be  necessary  to  reduce  the  two  equations 
in  two  unknowns  to  one  equation  in  one  unknown.    This 
process  is  called  elimination. 


374  GENEKAL  MATHEMATICS 

429.  Elimination  by  addition  or  subtraction.  The  two 
problems  which  follow  illustrate  the  method  of  elimination 
by  addition  or  subtraction. 

Solve  x  +  y  =  6,  (1) 

Solution.  Multiplying  (1)  by  3  so  as  to  make  the  coefficients  of 
//  numerically  the  same  in  both  equations, 

3  x  +  3  y  =  18  ("j) 

2ar-3y  =    2  (4) 

Adding,  or  =20 

Substituting  4  for  .c  in  (1),        4  +  ^  =  6. 
Solving  for  y,  y  =  2. 

4  +  2  =  6, 
Check.  8-6  =  2. 

This  method  is  called  elimination  by  addition.     Why  ? 

Solve  3;r-4y  =  l,  (1) 

2  a-  -f  5  y  =  —  7.  (2) 

Solution.  Multiplying  (1)  by  2  and  (2)  by  3  so  as  to  make  the 
coefficients  of  x  numerically  equal, 

6  x  -    8  y  =        2  (3) 

t;.,  -f  15 y  =  -21  (4) 

Subtracting,  -  23  y  =      23 

Substituting  —1  for  //  in  (1),  3  x  +  4  =1. 

3  a:  =-3. 

•    a 

-c  =—  1. 

Hefeee  y  =  —  1, 

and  x  =  —  1. 

This  method  is  called  elimination  by  subtraction.    Why  ? 


SIMULTANEOUS  LINEAR  EQUATIONS        375 

430.  Outline  of  elimination  by  addition  or  subtraction. 
To  solve  two  simultaneous  linear  equations  involving  two 
unknowns  by  the  method  of  addition  or  subtraction,  proceed 
as  follows  : 

1.  Multiply,  if  necessary,   the  members   of  the  first   and 
second  equations  by  such  numbers  as  irill  make  the  coefficients 
of  one  of  the  unknowns  numerically  the  same  in  both  equations. 

2.  If  the   coefficients   have   the    same   signs,   subtract   one 
equation  from  the  other;  if  they   have  opposite  signs,   add 
the  equations.    This  eliminates  one  unknown. 

3.  Solve  the  equation  resulting  from  step  2  for  the  unknown 
which  it  contains. 

4.  Substitute  the  value  of  the  unknown  found  in  step  3  in 
either  equation  containing  both  unknowns  and  solve  for  the 
second  unknown. 

5.  Check  the  solution  by  substituting  in  both  of  the  given 
equations  the  values  found. 

EXERCISES 

Solve  and  check  the  following  systems  by  the  method  of 
addition  or  subtraction  : 

'        3  x  +  2  y  —  7,             ox  —  3  '//  =  —  14,  3  x  —  o  //  =  23, 

-  2z  +  3y=8.         "  2a;  +  4//=10.  ''  1  x  +  <j  =  -  35. 

4z-3?/  =  -l,          7a-  +  9y=-lo,  2ar  +  3y  =  0. 

!'  5*  +  2y=16:       "  5ar-9y  =  -21.  ''  3x  +  2i/  =  o. 

llx-7y=-6,      .  :--2y  =  54 

' 


11.  x  =  y      y 


'  x  +  y  —  8. 


x         ?/  =    ,  m      n  _ 

•  +     = 


in  .  _ 

'•  $(,.+11,--  28.  3       10 


376 


GENERAL  MATHEMATICS 


60° 


FIG.  282 


GEOMETRIC  EXERCISES  FOR  ALGEBRAIC  SOLUTION 

13.  The  combined  height  of  a  tower  and  flagpole  is  110  ft. ; 
the  height  of  the  tower  is  70  ft.  more  than  the  length  of  the 
flagpole.    Find  the  height   of  the  tower  and  the  length  of 
the  flagpole. 

14.  A  rectangular  field  is  25  rd.  longer  than  it  is  wide.    The 
perimeter  of  the  field  is  130  rd.  Find  the  dimensions  of  the  field. 

15.  The  perimeter  of  a  football  field  is  320  yd.  and  the 
length   is   10  yd.    more   than   twice 

the  width.   What  are  its  dimensions  ? 

16.  The  circumference  of  a  circle 
exceeds  the  diameter  by  75  ft.    Find 
the  circumference  and  the  diameter. 
(Use  the  formula  C  =  -2y2  d.~) 

17.  Two  parallel  lines  are  cut  by 
a  transversal  forming  eight  angles, 

as  shown  in  Fig.  282.    Find  x  and  y,  and  all  of  the  angles. 

18.  The  interior  angles  on  the  same  side  of  a  transversal 
cutting  two  parallel  lines  are  (5  x  —  3  y)°  and  (x  +  y)°.    Their 
difference  is  70°.    Find  x  and  y. 

19.  Two  adjacent  angles  of  a  parallelogram  are  represented 

/o  \  o 

by    3  (2m  -  ri)°    and    i-g-  +  4  n }  , 

and  their  difference  is  30°.    Find  m,  n, 
and  the  angles  of  the  parallelogram. 

20.  The    difference    between    the 
acute  angles   of  a   right  triangle  is 

43°.    Find  the  acute  angles.  2g3 

21.  A    picture    frame    1  in.    wide 

(Fig.  283)  has  an  area  of  44  sq.  in.  The  picture  inside  the 
frame  is  4  in.  longer  than  it  is  wide.  Find  the  dimensions 
of  the  picture. 


SIMULTANEOUS  LINEAR  EQUATIONS        377 

431.  Elimination  by  the  method  of  substitution.     The 
following  problem  illustrates  the   method  of  elimination 
by  substitution : 

Solve  m  -f  n  =  5,  (1) 

2  m  -5n=-ll.  (2) 

Solution.    Solving  for  »i  in  terms  of  n  in  the  first  equation, 

w  =  5  -  n.  (3) 

Substituting  this  value  of  m  in  (2), 

2(o-i»)-5ti=-ll.  (4) 

10-2»»-5n=-ll. 
-  7  n  =  -  21. 

n-=.3.  . 

Substituting  3  for  n  in  (1),    m  +  3  =  5, 

m  =  2. 
Hence  m  =  22,  and  n  =  3. 

Check.  2  +  0  =  5, 

4  -  15  =  -  11. 

The  preceding  method  is  called  the  method  of  substitution. 

432.  Outline  of  steps  in  the  solution  of  a  system  by  the 
substitution  method  of  elimination.    To  solve  a  system  of 
linear  equations  containing  two  unknowns  by  the  method 
of  substitution,  proceed  as  follows: 

1.  Sol v<'  one  equation  for  one  unknown  in  terms  of  the  other. 

2.  Substitute  for  the  unknown  in  the  other  equation  the 
lvalue  found  for  it  in  step  1. 

3.  Solve  the  e<juation  resulting  in  step  2  for  the  unknown 
which  it  contains. 

4.  Substitute  the  value  of  the  unknown  obtained  in  step  3 
in  any  equation  containing  both  unknowns  and  solve  for  the 
second  unknown. 

5.  Check  the  solution  by  substituting  'in  the  given  equations. 


378  GENERAL  MATHEMATICS 

EXERCISES 

Solve  the  following  problems  by  the  method  of  substitution: 

3ff  +  46  =  ll,             -2ar-5?/?  =  17,  3m  +  2n  =  130, 

'  oa-b  =  3.  '  3x  +  2><;  =  2.  '  5 m  -  6 n  =  30. 

7s  +  3y  =  l,              2*  +  3y  =  8,  ^      y_ 

"llz-5y  =  21.  >'3z-5y  =  42.  5  +  2        ' 

9*  +  7»-=-12,           7ar-3y=8,  *'  x  _  // 

15*-2r=21            8»  +  2y  =  l34.  3      2 

NUMBER-RELATION*  PROBLEMS 

Solve  the  following  problems  by  the  method  of  substitution, 
and  check  by  one  of  the  other  methods : 

9.   Find  two  numbers  whose  sum  is  150  and  whose  difference 
is  10. 

10.  Find  two  numbers  whose  difference  is  15  such  that  when 
one  is  added  to  twice  the  other  the  sum  is  295. 

11.  Find  two  numbers  such  that  8  times  the  first  plus  4 
times  the  second  equals  100,  and  3  times  the  first  plus  7  times 
the  second  equals  87. 

12.  The  quotient  of  two  numbers  is  2  and  their  sum  is  54. 
Find  the  numbers. 

13.  The  value  of  a  certain  fraction  is  4-.   If  2  is  added  to  the 
numerator  and  7  to  the  denominator,  the  value  of  the  resulting 
fraction  is  ^.    Find  the  fraction. 

14.  The  sum  of  the  two  digits  in  a  two-place  number  is  8. 
If  18  be  subtracted  from  the  number,  the  resulting  number 
will  be  expressed  by  the  original  digits  in  reverse  order.    Find 
the  number. 

Solution.    Let  u  represent  the  digit  in  units'  place  and  t  the  digit 
in  tens'  place. 

Then  10  t  +  u  represents  the  original  number. 

From  the  first  condition,      t  +  u  =  8. 


SIMULTANEOUS  LINEAR  EQUATIONS        379 

From  the  second  condition, 

10 1  +  u  -  18  =  10  u  +  t.  (2) 

Simplifying  (2),  t  -  u  =  2.  (3) 

Solving  (1)  and  (3),  2  t  =  10. 

t  =  5. 

Substituting  5  for  /  in  (1),          «  =  3. 
Therefore  the  number  is  53. 

15.  The  tens'  digit  of  a  two-place  number  is  three  times 
the  units'   digit.     If  54  be  subtracted  from  the  number,  the 
difference  is  a  number  expressed  by  the  digits  in  the  reverse 
order.    Find  the  number. 

16.  If  a  two-digit  number  be  decreased  by  13,  and  this  dif- 
ference divided  by  the  sum  of  the  digits,  the  quotient  is  5. 
If  the  number  be  divided  by  one  fourth  of  the  units'  digit, 
the  quotient  is  49.    Find  the  number. 

433.  Summary  of  methods  of  elimination.  This  chapter 
has  taught  the  following  three  methods  of  solving  a  system 
of  simultaneous  linear  equations : 

1.  The  graphic  method. 

2.  Elimination  by  addition  or  subtraction. 

3.  Elimination  by  substitution. 

EXERCISES 

Some  of  your  classmates  may  be  interested  to  learn  a  fourth 
method  called  elimination  by  comparison.  Turn  to  a  standard 
algebra  and  report  to  your  classmates  on  this  method. 

MIXTURE  PROBLEMS 

Solve  the  following  problems  by  any  method : 
1.  A  grocer   has   two  kinds  of  coffee,  one  worth  300  per 
pound  and  another  worth  200  per  pound.    How  many  pounds 
of  the  30-cent  coffee  must  be  mixed  with  12  Ib.  of  the  20-cent 
coffee  to  make  a  mixture  worth  240  per  pound? 


380  GENERAL -MATHEMATICS 

2.  A  grocer  makes  a  mixture  of  20-cent  nuts  and  32-cent 
nuts  to  sell  at  28  £  a  pound.    What  quantities  of  each  grade  of 
nuts  must  he  take  to  make  60  Ib.  of  the  mixture  ? 

3.  How  much  milk  testing  5%  butter  fat  and  cream  testing 
25%  butter  fat  must  be  mixed  to  make  30  gal.  that  test  22% 
butter  fat  ? 

4.  How  much  milk  testing  3.7%  butter  fat  and  cream  testing 
25.5%  butter  fat  must  be  mixed  to  make  15  gal.  that  test  20% 
butter  fat  ? 

5.  "What  numl)er  of  ounces  of  gold  75%  pure  and  85%  pure 
must  be  mixed  to  give  10  oz.  of  gold  80%  pure  ? 

G.  An  alloy  of  copper  and  silver  weighing  50  oz.  contains 
5  oz.  of  copper.  How  much  silver  must  be  added  so  that  10  oz. 
of  the  new  alloy  may  contain  ^  oz.  of  copper  ? 

7.  The  standard  daily  diet  for  an  adult  requires  about  75  g. 
of  protein  and  100  g.   of  fat.     Mutton  (leg)  contains  19.8% 
protein  and  12.4%  fat.    Bread  (average)  contains  9.2%  protein 
and  1.3%  fat.    Find  how  many  grains  each  of  bread  and  mutton 
are  required  to  furnish  the  required  amount  of  protein  and  fat 
in  a  standard  ration  for  one  day. 

HINT.  Let  x  and  y  represent  the  number  of  grains  required  of 
mutton  and  bread  respectively. 

Then  0.198  x  +  0.092  y  =  75,  (1) 

and  0.121  x  +  0.013  y  =  100.  (2) 

Solve  the  equations  (1)  and  (2)  simultaneously. 

If  x  or  y  turns  out  negative,  we  know  that  it  is  not  possible  to  make 
up  a  standard  diet  out  of  the  two  foods  mentioned. 

8.  The   table   on  the    following   page    gives   the   amounts 
of  protein  and  fats  in  the  various  foods  often  used  in  the 
daily  diet. 


SIMULTANEOUS  LINEAR  EQUATIONS        381 


FOOD 

PER  CENT 
OF  PROTEIN 

PER  CENT 
OF  FAT 

Mutton  (le(r)    

19.8 

12.4 

Beef  (roast)     

22.3 

28.6 

Pork  (chops)    
Esfsrs    . 

16.6 
13.4 

30.1 
10.5 

Bread  (average)  

9.2 

1.3 

Beans  (dried)  
Cabbage  

22.5 
1.6 

1.8 
0.3 

Rice    

8.0 

0.3 

Find  three  pairs  of  food  combinations  that  will  make  a 
standard  diet  and  determine  the  number  of  grams  required  of 
each  in  the  following  list : 

(a)  Mutton  and  rice.  (f )  Bread  and  cabbage. 

(b)  Eggs  and  rice.  (g)  Beans  and  cabbage. 

(c)  Bread  and  eggs.  (h)  Bread  and  beans. 

(d)  Pork  and  bread.  ( i )  Beef  and  bread. 

(e)  Pork  and  beans.  (j)  Beef  and  rice. 

434.  Systems  of  equations  containing  fractions.  The  fol- 
lowing list  of  problems  offers  no  new  difficulties.  The 
student  merely  needs  to  remember  to  remove  the  fractions 
in  each  equation  by  multiplying  through  by  the  L.C.M.  of 
the  denominators  in  each,  thus  reducing  each  equation  to 
the  standard  form  ax  +  by  =  c,  where  a  represents  the  co- 
efficient of  x,  b  the  coefficient  of  y,  and  c  the  constant  term. 

5 
:6' 


3 

x  —  y 


or 


4  3         12 

The  first  equation  may  be  written 

x  +  y=i: 
Similarly,  (2)  reduces  to  x  —  y  =  \^ 


(1) 
(2) 


(3) 
Why? 


382  GENERAL  MATHEMATICS 

EXERCISES 
Reduce  to  the  standard  form,  and  solve : 


1. 

2. 
3. 

X  +  If 

a;  —  3  y 

m  +  1       1 

3 

2x  + 

3 

'/      3  ?/  4-  x 

> 

5 

4.   »  +  l       2' 
m-1       1 

3 
7  —  5 

4 

4 

1 

71-1            4 

«  -f  3      .s  +  4       5 

5 
•jr.  —  y 

3 

,  ^  +  y 

!«' 

2 

5.      5             10         2' 
f>  *       ~ 

2 

3 

./  i             ./ 

3 

Zt       .)        .s 

7           3 

0.3  x  +  0.7  y  =  5.7, 
2  x  —  4  y  =  —  4. 

3 

x  4-  5 

1                         !~ 

435.  Linear    systems    of    the    type    -+-  =  c\    work 

.     y 

problems.  In  the  problems  of  Art.  434  we  have  seen  the 
advisability  of  reducing  each  of  the  equations  in  a  system 
to  the  standard  form  by  eliminating  the  fractions  and 
collecting  similar  terms.  There  are  some  problems,  how- 
ever, in  which  it  is  advisable  to  solve  without  eliminating 
the  fractions.  An  example  will  illustrate  what  is  meant. 

Two  pipes  can  fill  ^  of  a  cistern  if  the  first  runs  2  hr. 
and  the  second  runs  3  hr.,  but  if  the  first  runs  3  hr.  and  the 
second  runs  2  hr.,  they  can  fill  ^  of  the  cistern. 

Solution.    Let  x  =  the  number  of  hours  it  will  take  the  first 

pipe  alone  to  fill  the  cistern, 
. 
and  y  =  the  number   of;    hours  it    will  take    the 

second  pipe  alone  to  fill  the  cistern. 

Then  -  =  the  part  of  the  cistern  the  first  pipe  can 

fill  in  1  hr. 

and  -  =  the  part  of  the  cistern  the  second  pipe 

y          can  fill  in  llir. 


SIMULTANEOUS  LINEAR  EQUATIONS        383 
From  the  first  condition. 

;  +  r^r  (1) 

From  the  second  condition, 

-  +  -  =  7!  (2) 

x      y      lo 

Multiplying  (1)  by  3  and  (2)  by  2, 
6      9  _  27 

and  -  +  -  =  ~  •  (4) 

x      y      15 

5      27      28 
Subtracting,  -  =  - —  •  (5) 

(Note  that  this  is  a  linear  equation  in  one  unknown.) 
Solving,  y  -  G, 

x  =  5. 
Problems  like  the  preceding  are  called  work  problems. 

WORK  PROBLEMS 

1.  A  and  B  can  build  a  fence  in  4  da.    If  A  works  6  da.  and 
quits,  B  can  finish  the  work  in  3  da.    In  how  many  days  can 
each  do  the  work  alone  ? 

2.  A  tank  can  be  filled  by  two  pipes  one  of  which  runs  3  hr. 
and  the  other  7  hr.,  or  by  the  same  two  pipes  if  one  runs  5  hr. 
and  the  other  6  hr.    How  long  will  it  take  each  pipe  alone  to 
fill  the  tank  ? 

3.  A  mechanic  and  an  apprentice  receive  $4.40  for  a  job  of 
work.     The  mechanic  works   5  hr.   and  the  apprentice   8  hr. 
Working  at  another  time,  and  at  the  same  rate  per  hour,  the 
mechanic  works  10  hr.   and  the  apprentice   11  hr.,  and  they 
receive  &7.30.    What  are  the  wages  per  hour  for  each? 


384  GENERAL  MATHEMATICS 

4.  Solve  the  following  problems  without  getting  rid  of  the 
fractions,  and  check  : 

1  1  _5  47_15 
H      —  7;  >  —    .  i 

*     x      y       6  3    x       y        4 

'J-i-i  'M  =  i. 

x       i/       (y  x      y 

2  3  =  19  11  _  9  =  26 

2     //i  n      15  ,      s        f        5 

'  A  .  ±  =  i  '  13_3  =  8 

m  n      5  s        t       5 

436.  Review  list  of  verbal  problems.  The  following 
problems  review  types  studied  in  earlier  chapters.  In  actual 
practice  many  problems  may  be  solved  by  using  either  one 
or  two  unknowns.  In  general  it  is  advisable  to  use  one  un- 
known, but  sometimes  it  is  easier  to  translate  the  problem 
into  algebraic  language  if  two  unknowns  are  used.  It  will 
be  helpful  if  some  member  of  the  class  will  show  the  two 
methods  in  contrast. 

MOTION  PROBLEMS 

1.  A  crew  can  row  8  mi.  downstream  in  40  min.  and  12  mi. 
upstream  in  1  hr.  and  30  min.  Find  the  rate  in  miles  per  hour 
of  the  current  and  the  rate  of  the  crew  in  still  water. 

Solution.    Let  x  =•  the  rate  of  the  crew  in  still  water, 

and  y  =  the  rate  of  the  current. 

Then,  if  we  express  the  rates  in  miles  per  hour, 


x-y  =        =  S.  (2) 

(TO 

Adding,  2  x  =  20. 

Hence  x  =  10,  the  rate  of  the  crew  in  still  water, 

and  y  =  2,  the  rate  of  the  current. 


SIMULTANEOUS  LINEAR  EQUATIONS        880 

2.  A  boatman  rowed  10  mi.  upstream  and  4  mi.  back   in 
3'  lir.    If  the  velocity  of  the  current  was  2  mi.  per  hour,  what 
was  his  rate  of  rowing  ? 

3.  The  report  of  a  gun  traveled  367yd.  per  sec.  with  tin- 
wind  and  353  yd.  per  sec.  against  the  wind.    Find  the  velocity 
of  sound  and  the  rate  at  which  the  wind  was  blowing. 

4.  An  ae'roplane  flying  with  a  wind  blowing  at  the  rate  of 
2  mi.  an  hour  consumes  2  hr.  30  min.  in  going  a  certain  dis- 
tance  and  2  hr.  42  min.  30  sec.  in  returning.    Find  the  distance 
and  the  rate  of  the  aeroplane  in  still  air. 

BEAM  PROBLEMS 

NOTK.  Beam,  or  lever,  problems  which  involve  two  unknowns 
are  readily  solved  by  means  of  the  laws  of  leverages  and  forces 
discussed  in  Art.  233. 


5.  Two  boys  carry 
a  bag  of  coal  weigh- 
ing 25  Ib.  by  hanging  M 


it  on  a  pole  8  ft.  long     k  I       n,__  J~ 

at   a   point   2  ft.    from      I 

the  middle  of  the  pole.     +x  JJ5  +y 

How  much  of  the  -load  F      284 

does  each  boy  carry  ? 

Solution.    Let  x  and  y  represent  the  number  of  pounds  carried 
by  each  boy. 

Then,  using  M,  Fig.  284  (the  middle  of  the  pole),  as  the  turning 
point, 

4  y  -  50  -  4  x  =  0.    (Why  ?) 

From  which  //  —  x  =  12. .1.  (1) 

x  +  y  =  25.      (Why  ?)  (2) 

Adding  (1)  and  (2),  2  ?/  =  37.f>. 

r~i8f. 

Substituting  for  y  in  (2),     18;}  ./•  =  (i\. 


386  GENERAL  MATHEMATICS 

6.  Two  weights  balance  when  one  is  14  in.  and  the  other 
10  in.  from  the  fulcrum.  If  the  first  weight  is  increased  by 
2  Ib.  and  placed  10  in.  from  the  fulcrum,  the  balance  is  main- 
tained. Find  the  number  of  pounds  in  each  weight. 
"  --3«Two  weights  balance  when  one  is  12  in.  and  the  other 
is  10  inT^em  the  fulcrum.  The  balance  is  maintained  if  the 
first  weight  isriKived  3  in.  nearer  the  fulcrum  and  if  3  Ib.  is 
subtracted  from  the  s&soml.  Find  the  weights. 

8.  An  iron  bar  6  ft.  loug^^^ighing  20  Ib.  is  used  by  two 
boys,  one  at  each  end,  to  carry  a  Io9ji-oi^oO  Ib.    How  many 
pounds  must  each  boy  carry  if  the  load  hangs~2~ttr-feaio_the 
right  end  ?    (Consider  the  weight  of  the  entire  bar  as  hanging 
at  the  middle  of  the  bar.) 

9.  A  wagon  bed  12  ft.  long  is  loaded  with  20  cakes  of  ice 
weighing  250  Ib.    The  bed  extends  2  ft.  over  the  front  axle  of 
the  running  gears  and  3  ft.  behind  the  rear  axle.   Find  the  load 
supported  by  each  axle. 

10.  The  material  in  a  30-foot  bridge  weighs  3600  Ib.    The 
bridge  supports  two  loads :   700  Ib.  at  3  ft.  from  one  end,  and 
1500  Ib.  at  5  ft.  from  the  other  end.    Find  the  loads  borne  by 
the  two  supports. 

11.  Three  men  have  to  carry  an  oak  beam  15  ft.  long  weigh- 
ing 250  Ib.    Two  of  the  men  lift  at  the  ends  of  an  iron  bar 
placed  crosswise  beneath  the  beam,  and  the  third  man  lifts  at 
the  rear  end  of  the  beam.    Where  must  the  iron  bar  be  placed 
in  order  that  each  man  will  carry  one  third  of  the  load  ? 

*  RECREATION  PROBLEMS 

12.  A   man  upon   being  asked  the   age  of   his  two  sons, 
answered,  "  If  to  the  sum  of  their  ages  18  be  added,  the  result 
will  be  double  the  age  of  the  elder ;  but  if  6  be  taken  from  the 
difference  of  their  ages,  the  remainder  will  be  equal  to  the  age 
of  the  younger."    Find  the  age  of  each. 


SIMULTANEOUS  LINEAR  EQUATIONS        387 

13.  In  a  guessing  game  the  leader  says,  "  If  you  will  add 
10  years  to  your  age,  divide  the  sum  by  your  age,  add  6  to  the 
quotient,  and  tell  me  the  result,  I  will  tell  you  your  age."    How 
did  he  find  it  ? 

14.  A  baseball  team  has  played  40  games,  of  which  it  has 
won  28.    How  many  games  must  it  win  in  succession  in  order 
to  bring  its  average  of  games  won  up  to  0.750  ? 

15.  A  girl  has  worked  12  problems.    If  she  should  work  13 
more  problems  and  get  8  of  them  right,  her  average  would  be 
72%.    How  many  problems  has  she  worked  correctly  thus  far  ? 

16.  Two  bicycle  riders  ride  together  around  a  circular  track, 
one  along  the  outside  edge,  where  the  radius  of  the  circle  is 
R,  and  the  other  along  the  inside  edge,  where  the  radius  is  r. 
One  revolution  of  the  pedals  carries  the  former's  bicycle  20  ft. 
and  the  latter's  25  ft.    Write  a  formula  expressing  the  differ- 
ence between  the  number  of  pedal  revolutions  made  by  the  two 
cyclists  in  going  around  the  track  once ;  five  times  ;  n  times. 

17.  If  10  marbles  of  one  size  are  dropped  into  a  bucket  of 
water,  and  the  water  rises  a,  inches,  and  15  equal  marbles  of 
another  size  are  dropped  into  the  same  bucket,  and  the  water 
rises  b  inches,  write  a  formula  showing  how  many  times  larger 
one  of  the  first  size  is  than  one  of  the  second  size. 

18.  An  automobile  .tire  when  fully  inflated  has  a  radius  of 
18  in.  Owing  to  a  leakage  of  air,  this  is  reduced  to  17  in.    Indi- 
cate how  many  more  revolutions  per  mile  are  necessary  because 
of  the  leakage.    If  the  original  radius  is  R  and  the  reduced 
radius  r,  what  is  the  formula  which  could  be  used  to  calculate 
the  difference  of  revolutions  per  mile  ? 

19.  Divide  $183  into  two  parts,  so  that  ^  of  the  first  part 
shall  be  equal  to  y3^  of  the  second  part. 

20.  Each  of  two  brothers  wanted  to  buy  a  lot  valued  at 
$240.    The  elder  brother  said  to  the  younger,  "  You  lend  me 
|  of  your  money,  and  I  can  purchase  the  l°t-"    "But,"  said  the 


088  GENERAL  MATHEMATICS 

younger  brother,  ''you  lend  me  |  of  your  money,  and  I  ran 
purchase  the  lot."    How  much  money  did  each  have '.' 

21.  A  group  of  boys  bought  a  touring  car.    After  paying  for 
it  they  discovered  that  if  there  had  been  one  boy  more,  They 
would  have  paid  $30  apiece  less,  but  if  there  had  been  our 
boy  less,  they  would  have  paid  $60  apiece  more.    How  many 
boys  were  there,  and  what  did  they  pay  for  the  car? 

22.  The  Champion  American  League  team  one  year  (1914) 
won  46  more  games  than  it  lost.    The  team  standing  second 
played  153  games,  winning  8  less  than  the  first  and  losing 
9  more  than  the  first.    Find  the  number  of  games  won  and 
lost  by  each  team. 

23.  It  is  said  that  the  following  problem  was  assigned  by 
Euclid  to  his  pupils  about  three  centuries  B.C.  :  "  A  mule  and 
a  donkey  were  going  to  market  laden  with  wheat.    The  mule 
said  to  the  donkey,  '  If  you  were  to  give  me   one  measure, 
I  would  carry  twice  as  much  as  you ;  if  I  were  to  give  you  one 
measure,  our  burdens  would:  be  equal.'  What  was  the  burden 
of  each  ?  " 

*  MISCELLANEOUS  PROBLEMS 

24.  A  bar  30  in.  long  is  balanced  by  a  40-pound  weight  at  one 
end  and  a  32-pound  weight  at  the  other  end.    Find  the  position 
nl'  the  support. 

25.  A  man  has  a  f 2000  exemption  frOm  income  tax,  but  pays 
at  the  rate  of  6%  on  the  rest  of  his  income.    He  finds  that  after 
paying  income  tax  his  actual  income  is  $3410.   On  what  amount 
does  he  pay  the  6%  tax  ? 

26.  A  chemist  has  the  same  acid  in  two  strengths.    If  16  1. 
of  the  second  are  mixed  with  241.  of  the  first,  "the  mixture  is 
42%  pure,  and  if  61.  of  the  first  are  mixed  with  41.  of  the 
second,  the  mixture  is  43%  pure.    Find  the  per  cent  of  purity 
of  each  acid.    Problems  like  this  are  often  given  as  practical 
problems.    Find  out  from  some  chemist  why  it  is  not  practical. 


SIMULTANEOUS  LINEAR  EQUATIONS        389 

27.  After  a  strike  a  corporation  decided  to  raise  the  wages 
of  each  laborer  from  .?•  to  y  by  the  formula  y  =  tux  +  i>,  where 
ra  and  b  are  to  be  determined  by  the  facts  that  a  man  who 
made  $2  is  to  receive  $2.30,.  and  one  who  made  $2.20  is  to 
receive  $2.41.    Find  ra  and  b,  also  the  new  wage  of  a  man 
who  formerly  received  $3,  $4,  $4.20. 

28.  At  what  market  price  must  one-year  5%  bonds  be  offered 
for  sale  in  order  that  the  buyer,  by  holding  them  until  maturity, 
nitiy  make  6%  on  his  investment? 

HINT.  The  profit  made  must  come  from  two  sources  :  the  interest 
on  the  par  value,  which  is  $5,  and  the  excess  of  the  maturity  value 
over  the  price  paid  for  the  bonds.  If  x  is  the  price  paid,  then 
100  -  x  +  5  =  0.06  x. 

SUMMARY      - 

437.  This  chapter  has  taught  the  meaning  of  the  fol- 
lowing words  and  phrases :  simultaneous  equations,  linear 
systems  of  equations,  indeterminate  equations,  contradic- 
tory equations,  identical  equations,  elimination. 

438.  This  chapter  has  taught  the  following  methods  of 
solving  a  system  of  equations  in  two  unknowns: 

1.  Solution  by  graph. 

2.  Solution  by  addition  or  subtraction. 

3.  Solution  by  substitution. 

439.  The  student  has  been  taught  how  to  solve  systems 

involving  fractions,  and  systems  of  the  type  -  -f  -  =  c. 

x     y 

440.  The  following  types  of  verbal  problems  have  been 
introduced:  geometric  problems,  number-relation  problems, 
mixture  problems,  work  problems,  motion  problems,  beam 
problems,  and  recreation  problems. 


CHAPTER  XVI 

GEOMETRIC  AND  ALGEBRAIC  INTERPRETATION  OF 
ROOTS  AND  POWERS 

441.  Introductory  work;    square    root.     The   following 
exercises  are  introductory  to  the  work  of  the  chapter. 

EXERCISES 

1.  What  number  multiplied  by  itself  equals  9?  16?  121? 
169?  x2?  if? 

2.  How  many  answers  are  there  to  each  part  of  Ex.  1? 
(Why?) 

3.  One  of  the  two  equal  factors  of  a  number  is  called  the 
square  root  of  the  number.    What  is  the  square  root  of  49  ? 

4  4y2 

of  64?  of  625?  ofar2?  of  4^?  of  J?  o£^-J 

9          9y* 

4.  The  positive  square  root  of  a  number  is  indicated  by  a 
sign  ( V     )  called  the  radical  sign,  and  either  the  radical  sign 
alone  (V     )  01>  the   radical  sign  preceded  by  the  plus  sign 
(+  V     )  means  the  positive  square  root  of  the  number  under- 
neath the  sign.     The  number  underneath  the  radical  sign  is 
called  the  radicand.    The  negative  square  root  is  indicated  by 
the  radical  sign  preceded  by  the  minus  sign  (—  V     )•    With 
the  preceding  definitions  in  mind  give  the  value  of  the  fol- 
lowing: V25;   Vl6;  VlOO;  -Vl21;  V(X25 ;  -Vl44;  Vj; 


9    '    \25iy2 

5.  Express  the  following  statement  by  means  of  a  formula: 
A  number  y  equals  the  square  of  another  number  x. 

390 


INTERPRETATION  OF  ROOTS  AND  POWERS    391 

6.  If  x  =  1  in  the  formula  y  =  ar2,  what  is  the  value  of  y? 
If  x  =  2,  what  is  the  value  of  y  ? 

7.  Calculate  the  corresponding  values  of  y  in  the  formula 
y  =  x'  for  each  of  the  following  values  ofic:  —  1 ;   —  2  ;   -(-3; 

Q.      i     1  .      i     2  .  2 

~~   t>  y     ~7~  7)~"  j     T  "3*  5      "*"""    Q  • 

8.  Fill    in  the   proper   values    in    the   following   table   of 
squares  and  square  roots  for  use  in  the  next  article. 


X 

i 

-  1 

2 

-  2 

±3 

£4      ± 

5      ±6 

b 

± 

I 

i 

i 

y 

i 

i 

i 

4 

. 

i 

X 

±4 

± 

T 

t 

s  ± 

i    i 

:  j 

-     ±i 

±1      ± 

V 

t 

\ 

i 

±- 

¥ 

y 

i. 

-I    ^ 

^ 

> 

y 

r 

~J 

i 

J 

\ 

/ 

\ 

/ 

j 

\L 

/ 

L 

1 

V 

^ 

\ 

^ 

A 

/ 

m 

J 

^ 

/ 

1  ^ 

^  k 

^i 

\^ 

.  ' 

"s^  L 

* 

^  ^ 

^^ 

tSi 

^  ^ 

-  JF 

"%i.  - 

„ 

5  J 

•    'T 

"  ~  5  " 

n  ~ 

-' 

r-i-«  p 

i; 

-, 

5 

-:t~ 

±t 

_t  —  : 

i^-^ 

f" 

FIG.  285.   DEVICE  FOR  FINDING  SQUARES  AND  SQUARE  ROOTS 

442.  Graph  of  y  =  x*\  a  device  for  finding  squares  and 
square  roots.  The  values  of  the  preceding  table  have 
been  plotted  in  Fig.  285.  Values  for  x  were  laid  off 


392  GENERAL  MATHEMATICS 

horizontally  on  the  2>axis,  and  the  corresponding  values 
for  y  vertically  on  the  ^-axis.  The  points  were  then  con- 
nected by  a  smooth  curve,  as  shown.  This  curve  serves 
as,  a  device  for  determining  squares  and  square  roots,  as 
we  shall  now  see. 


EXERCISES 


1.  Determine  by  the  graph  in  Fig.  285  the  square  root  of  1 »'» : 
9;  2,0;  22;  3;  2.    How  many  answers' do  you  obtain  for  the 
square  root  of  9  ?  4  ?  25  '.'  for  the  square  root  of  each  number 
shown? 

2.  By  means  of  the  graph  in  Fig.  285  find  the  square  of  2; 
1.4;   2.2;   2.4;  3.3;  5.6;  3.9;   1.7. 

3.  Check  your  results  for  Ex.  2  by  squaring  the  numbers 
given.    The  squares  should  be  approximately  those  you  found 
by  means  of  the  graph. 

4.  How  would  you  make  a    graph  that  would    give   you 
squares  and  square  roots  more  accurately  than  the  graph  in 
Fig.  285  ? 

443.  A  positive  number  has  two  square  roots.   The  graph 
shows  that  the  square  root  of  4  equals  either  -j-  *2  or  —  2  ; 
that  is,  there  are  two  answers  for  the  square  root  of  a 
positive    number.    Thus,    (—3)  (—3)  =9,    as    also    does 
(  -H  3)  ('+  3  ).    Note  the  symmetry  of  the  curve  in  Fig.  285 
and  see  if  you  can  explain  it. 

444.  Quadratic  surd.     The   indicated  square  root  of  a 
number  which  is  not  a  perfect  square  is  called  a  quadratic 

*« i\l :  for  example,  v3,  v20,  V.r. 

« 

445.  Quadratic  trinomial.    Trinomials  like  a2  -f  2  ab  +  b2 
and  .?-'2  —  2  .ry  +  y2  are  of  the  second  degree,  and  are  called 
ijuadratic  trinomials.    The  word  "  quadratic  "  comes  from 


INTERPRETATION  OF  ROOTS  AND  POWERS    393 


the  old  word  "  quadrature,"  which  means  a  geometric 
square  ;  hence  quadratic  means  that  a2  4-  2  ab  -f  52  and 
./2  —'2xy  +  y2  are  of  the  second  degree.  They  are  the 


a 
ab 

b         b2 

«2 

ab 

ab 


ab 


a       +      b 

FIG.  286.    GEOMETRIC  REPRESENTATION  OF  (a  +  b)2. 

squares  of  a  +  b  and  ^  —  y  respectively,  as  we  have  already 
seen  earlier  in  this  book.  In  Fig.  286  the  geometric  square 
of  a  +  b  is  shown. 

EXERCISES 

1.  See  if  you  can  point  out  where  we  have  worked  with 
problems   dealing   with   the   square   of   binomials    like   a  +  b 
and  x  —  y. 

2.  Where,  earlier  in  the  book,  have  we  learned  how  to  take 
the  square  root  of  trinomial  squares  by  factoring  ? 

3.  Find  the  value  of  the  following  : 


Vra2  +  2  mn  +  n*  ;  V4  "a?  +  12  xy  +  9  if  ;  V9  a2  -  12  «6  +T7X 

*446.  Square  roots  of  algebraic  polynomials  and  arith- 
metical numbers.  We  have  already  seen  in  Chapter  IX 
how  the  square  root  of  a  trinomial  is  found  by  reversing  the 
process  of  squaring  a  binomial  :  that  is,  .by  finding  one  of 
the  two  equal  factors  of  the  trinomial.  The  same  process 
may  be  explained  geometrically  if  the  exercises  given  on 
page  394  are  carefully  solved. 


394 


GENERAL  MATHEMATICS 


ab 


ab 


FIG.  287 


ILLUSTRATIVE  EXERCISES 

1.  Find  the  square  root  of  16  x~  +  40a-y  +  25  y2 

Solution.  If  this  trinomial  is  a  perfect  square  of  some  binomial,  it 
may  be  illustrated  by  Fig.  287,  in  which  the  side  of  the  largest  square 
obtained  by  inspection  and  corresponding  to  a2  is  4  x.  Therefore 
the  side  of  each  rectangle  corresponding  to  , 

each  ab  is  4  x,  and  the  area  correspond- 
ing to  2  ab  +  b2  must  be  40  xy  +  25  y'2.  The 
problem  therefore  consists  in  determining 
the  width  of  the  strip  which  we  are  adding 
on  two  sides  and  which  corresponds  to  the 
b  of  the  formula.  In  this  case  b  is  5  y.  Now 
5  y  may  be  obtained  by  dividing  40  xy  by 
the  sum  of  4  x  and  4  x,  or  8  x.  Hence 
doubling  the  term  already  found  (4  x)  the 
result  8  x  serves  as  a  divisor  for  determin- 
ing the  next  term.  Fig.  287  shows  that  we  double  4  x  because 
8#  is  approximately  the  combined  length  of  the  strip  to  which 
we  are  adding.  This  is  illustrated  more  clearly  in  the  next  problem. 

2.  Two  boys   were  asked  to    stake   out  a  square    plot  of 
ground  with   an   area  .of    4225  sq.  ft.  5 
What  is  the  length  of  a  side  ? 

Solution.  The  boys'  thinking  about  the 
problem  might  take  some  such  form  as 
follows : 

(a)  It  is  obvious  that  we  can  make  it 

at  least  60  by  60.    We  shall  suppose  that  60 

this   is  constructed.     See  the  square  with 

unbroken  lines  (Fig.  288).    This  uses  up  pIGi  288 

:5600  sq.  ft.,  leaving  625  sq.  ft.   We  can  add 

to  the  square  already  constructed  by  adding  to  two  sides  and  still 

keep  it  a  square. 

(b)  The  combined  length  of  the  edges  to  which  we  are  adding 
is    120  ft.     Hence  the   approximate    length   of  the   strip  added   is 
120  ft.    Why  approximate? 

(c)  120  is  contained  five  times  in  625  (with  a  remainder). 


INTERPRETATION  OF  ROOTS  AND  POWERS    395 


(d)  If  we  make  the  strip  5  ft.  wide,  the  total  length  will  be  125 
(for  one  strip  will  be  60  ft.  and  the  other  65  ft.). 

(e)  125  is  contained  exactly  five  times  in  625. 

(f)  Hence  the  square  must  be  constructed  so  as  to  be  65'  by  65'; 
that  is,  the  square  root  of  4225  is  65. 

3.  Find  the  square  root  of  the   polynomial  a2  +  2  ab  +  b2 
+  2ac  +  2bc  +  c2  (see  Fig.  289). 

Solution.  The  side  of  the  largest  square  is  a^  therefore  the  trial 
divisor  is  2  a.  The  width  of  the  first  strip  is  b,  therefore  the  divisor 
is  2  a  +  b.  Multiplying  by  b  and  subtracting  the 
remainder  gives  2  OK  +  2  be  +  c2.  The  length  of 
the  square  now  constructed  is  a  +  b.  The  edge 
to  which  we  are  adding  is  2  a  +  2  b  units  long 
(trial  divisor).  2  a  +  2  b  is  contained  c  times 
in  2  ac  +  2  be.  If  we  make  the  strip  c  units 
wide,  the  total  length  of  the  strip  to  which  we 
add  is  2  a  +  2  b  +  c  (complete  divisor).  (Why  ?) 
Multiplying  and  subtracting,  the  remainder  is 
zero.  The  side  of  the  total  square  is  a '+  b  +  c,  or 


FIG.  289 


Va2  +  2  ab  +  bz  +  2  ac  +  2  be  +  c2  =  a  +  b  +  c. 
The  work  may  be  arranged  as  follows : 
Largest  square,  a2  a2  +  2  ah  +  b2  +  2  ac  +  2  be  4-  c2  [  a  +  b  +  c 


2  ab  +  b* 

2  ah  +  b2 


First  trial  divisor,  2  a 
First  complete  divisor,  2  <i  + l> 
Second  trial  divisor,  2  a  +  2  b 
Second  complete  divisor,  2  a  +  2  I  +  c 


2  ac  +  2  be  +  c2 
2  ac  +  2  be  +  c2 


4.  Find  the  first  digit  in  the  square  root  of  177,2^1. 

Solution.  To  determine  this  first  digit  we  must  remember  (a)  that 
the  square  of  a  number  of  one  digit  consists  of  one  or  two  digits,  the 
square  of  a  number  of  two  digits  consists  of  three  or  four  digits,  the 
square  of  a  number  of  three  digits  consists  of  five  or  six  digits,  and  so 
on ;  (b)  that  the  number  of  digits  in  the  integral  part  of  the  square 
of  a  number  is  twice  as  large  or  one  less  than  twice  as  large  as  the 
number  of  digits  in  the  integral  part  of  the  given  number.  This 
suggests  the  following  device  for  determining  the  number  of  digits 


396  GENEKAL  MATHEMATICS 

in  the  integral  part  of  the  square  root  of  a  number.  Beginning  at 
the  decimal  jioint,  mark  off  toward  the  left  groups  of  two  digits 
each.  Then  the  number  of  digits  in  the  square  root  will  be  the 
same  as  the  number  of  groups.  Thus,  since  177,241  is  made  up  of 
three  groups  of  two  digits  (17'72'41'),  the  square  root  of  177,241 
contains  three  digits  in  its  integral  part.  We  are  thus  able  to  esti- 
mate the  largest  square  as  400  (that  is,  the  first  digit  is  4)  and  then 
proceed  as  in  Ex.  3.  The  work  may  be  arranged  as  follows : 

17  72'41 1 400  +  20  +  1 

16  00  00 

First  trial  divisor,  800  I  1  72  41 
First  complete  divisor,  820 1  1  64  00 
Second  trial  divisor,  840 
Second  complete  divisor,  841 
Therefore  Vl77241  =  421. 

*447.  Steps  involved  in  finding  square  roots.  The  follow- 
ing steps  were  used  in  Exs.  1~4,  above  ;  the  student  should 
study  them  carefully. 

1.  Estimate  the  largest  square  in  the  number. 

2.  Double  the  root  already  found  for  a  trial  divisor. 

3.  Divide   the  first   term   of  the   remainder   by   the   trial 
ilirt'xor,  placing  the  quotient  as  the  next  term  of  the  root-. 

4.  Annex  the  term  just  found  to  the  trial  divisor  to  form  a 
complete  divisor  and  continue  the  process  until  the  other  fcn/i* 
f>f  the  root  are  found. 

EXERCISES 

Find  the  square  roots  of  the  following  polynomials : 

1.  <>'2  +  2ab  +  b2.  -"3.  4«4+-4«8  +  9a2  +  4  0+4. 

2.  16z*  +  2±xy  +  9//'.  6.  x4  -  2x3  +  3x2  -  2  x  +  1. 

3.  49  if  -  14  yz  +  z\  7.  1  -  4  a  +  6  «.2  -  4  «8  +  «\ 
—4.  x*+2x8  +  3x2  +  2r  +  l.      8. 

9.  x6  +  4  ax*  -  2  aV  +  4  aV  -  4  <>\r  +  a6. 
10.  9  +  12  //  +  6  f  +  4  ?/  +•  4  if  +  if. 


INTERPRETATION  OF  ROOTS  AND  POWERS    397 


11.  **  +  8. 

,   '.9, 

12.  7  +  6a 
4 

13.  576. 

14.  9025. 

15.  51,529. 


_  16       4  16.  61,504. 

~  x       42'         17.  57,121. 
6  a*  2_8a3         18.  2. 

«£2  #2      .      NOTE.    Write  2.00WOO  and 

proceed  as  in  Ex.  4,  Art.  446. 

19.  3. 

20.  3.1416. 


448.  Table  of  roots  and  powers.    In  practical  situations 
it  is  convenient  to  use  a  table  of  roots  and  powers.    There 
are  a  number  of  very  useful  tables  in  textbook  or  leaflet 
form,  and  the  student  is  now  in  a  position  where  he  c,an 
easily  learn  how  to  use  them.    A  very  simple  table   of 
roots  and  powers  is  submitted  on  page  398.    It  will  fre-- 
quently  prove  a  great  convenience  to  the  student  in  his 
work  on  the  following  pages. 

449.  The  theorem  of  Pythagoras.   If  we  study  the  follow- 
ing  exercise    carefully   we    shall    discover    a   well-known 
geometric  theorem  which  will  be  useful  in  later  work. 

EXERCISE 

Construct  a  right  triangle,  making  the  'sides  including  the 
right  angle  3  and  4  units  long  respectively  (see  AAJ'>( ',  Fig.  290). 
Using  the  same  unit,  find  the  length  of  J  />'. 
On  each  side  draw  a  square  and  divide 
each  square  into  unit  squares.  Counting 
these  squares,  find  how  the  square  on  the 
hypotenuse  compares  with  the  sum  of  the 
squares  on  the  other  two  sides. 

The   preceding    exercise    illustrates 

,.    ,  Fro.  290 

the  familiar  theorem  of   Pythagoras: 

///  it  right  triangle  the  sum  of  the  squares  on  the  sides  inclml- 
l n < i  the- -right  angle  is-^ual  to  the  .squaw  on.  tlie  '• 


398 


GENERAL  MATHEMATICS 


TABLE  OF  ROOTS  AND  POWKKS 


No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

1 

1 

1 

1.000 

1.000 

51 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.259 

52 

2,704 

140,608 

7.211 

3.732 

3 

9 

27 

1.732 

1.442 

53 

2,809 

148,877 

7.280 

3.756 

4 

16 

64 

2.000 

1.587 

54 

2,916. 

157,464 

7.348 

3.779 

5 

25 

125 

2.236 

1.709 

55 

3,025 

166,375 

7.416 

3.802 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.825 

7 

49 

343 

2.645 

1.912 

57 

3,249 

185,193 

7.549 

3.848 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

7.615 

3.870 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

7.681 

3.892 

10 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

7.745 

3.914 

11 

121 

1,331 

3.316 

2.223 

61 

3,721 

226,981 

7.810 

3.936 

12 

144 

1,728 

3.464 

2.289 

62 

3,844 

238,328 

7.874 

3.957 

13 

169 

2,197 

3.605 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

1% 

2,744 

3.741 

2.410 

64 

4,096 

262,144 

8.000 

4.000 

15 

225 

3,375 

3.872 

2.466 

65 

4,225 

274,625 

8.062 

4.020 

16 

256 

4,096 

4.000 

2.519 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.061 

18 

324 

5,832 

4.242 

2.620 

68  • 

4,624 

314,432 

8.246 

4.081 

19 

361 

6,859 

4.358 

2.668 

69 

4,761 

328,509 

8.306 

4.101 

20 

400 

8,000 

4.472 

2.714 

70 

4,900 

343,000 

8.366 

4.121 

21 

441 

9,261 

4.582 

2.758 

71 

5,041 

357,911 

8.426 

4.140 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.795 

2.843 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,824 

4.898 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

5.000 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

5,776 

438,976 

8.717 

4.235 

27 

729 

19,683 

5.196 

3.000 

77 

5,929 

456,533 

8.774 

4.254 

28 

784 

21,952 

5.291 

3.036 

78 

6,084 

474,552 

8.831 

4.272 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.290 

30 

900 

27,000 

5.477 

3.107 

80 

6,400 

512,000 

8.944 

4.308 

31 

961 

29,791 

5.567 

3.141 

81 

6,561 

531,441 

9.000 

4.326 

32 

1,024 

32,768 

5.656 

3.174 

82 

6,724 

551,368 

9.055 

4.344 

33 

1,089 

35,937 

5.744 

3.207 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

5.830 

3.239 

84 

7,056 

592,704 

9.165 

4.379 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.219 

4.396 

36 

1,296 

46,656 

6.000 

3.301 

86 

7,396 

636,056 

9.273 

4.414 

37 

1,369 

50,653 

6.082 

3.332 

87 

7,569 

658,503 

9.327 

4.431 

38 

1,444 

54,872 

6.164 

3.361 

88 

7,744 

681,472 

9.380 

4.447 

39 

1,521 

59,319 

6.244 

3.391 

89 

7,921 

704,969 

9.433 

4.464 

40 

1,600 

64,000 

6.324 

3.419 

90 

8,100 

729,000 

9.486 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.497 

42 

1,764 

74,088 

6.480 

3.476 

92 

8,464 

778,688 

9.591 

4.514 

43 

1,849 

79,507 

6.557 

3.503 

93 

8,649 

804,357 

9.643 

4.530 

44 

1,936 

85,184 

6.633 

3.530 

94 

8,836 

830,584 

9.695 

4.546 

45 

2,025 

91.125 

6.708 

3.556 

95 

9,025 

857,375 

9.746 

4.562 

46 

2,116 

97.336 

6.782 

3.583 

96 

9,216 

884,736 

9.797 

4.578 

47 

2,209 

103,823 

6.855 

3.608 

97 

9,409 

912,673 

9.848 

4.594 

48 

2,304 

110,592 

6.928 

3.634 

98 

9,604' 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7.000 

3.659 

99 

9,801 

970,299 

9.949 

4.626 

50 

2,500 

125,000 

7.071 

3.684 

100 

10,000 

1,000,000 

10.000 

4.641 

INTERPRETATION  OF  ROOTS  AND  POWERS    399 

This  theorem  is  one  of  the  most  famous  theorems  of 
geometry.  Centuries  before  Christ  the  Egyptians  used  a 
rope  divided  by  knots  so  that  its  three  lengths  were  in 
the  ratio  3:4:5.  This  rope  was  used  in  land  surveying 
and  also  in  the  orientation  of  their  temples.  In  fact, 
we  read  of  professional  "  rope  fasteners "  (surveyors  ?). 
Furthermore,  the  proof  of  the  theorem  itself  has  always 
appealed  to  the  interest  of  mathematicians.  When  we  shall 
have  advanced  in  our  study  of  mathematics  it  will  be 
possible  for  the  student  to  find  many  proofs  of  this  theorem 
that  he  can  understand.  The  earliest  general  proof  is 
credited  to  Pythagoras,  who  lived  about  500  B.C. 

The  student  has  probably  found  this  theorem  to  be  the 
basis  for  one  of  the  most  useful  rules  of  arithmetic.  The 
proof  given  in  arithmetic  classes  is  usually  that  given  in 
the  exercise  above.  However,  a  general  proof  demands 
that  we  prove  the  theorem  independent  of  the  accuracy 
of  the  figure  (that  is,  independent  of  the  measurements 
and  constructions  involved).  We  shall  presently  give 
such  a  proof.  The  exercises  which  follow  are  intended 
to  review  the  material  necessary  to  establish  this  proof. 

EXERCISES 

1.  In  Fig.  291 A  A  EC  is  a  right  triangle,  right-angled  at  C,  with 
CD±  AB.  Re  view  the  proof  which  shows  that  AADC^  A  ABC. 

c       b 

2.  Prove  that  in  Fig.  291  -  =  - 

and  that  b2  =  cm. 

b; 

3.  Review  the  proof  which  shows 

that  ABDC-^  A  ABC.  A\ 

4.  Prove  that  in  Fig.  291  -  =  - 

O  n  41 


5.  Show  by  using  Exs.  2  and  4  that  a2  +  tf  =  <?. 


400 


( i  ION  EKAL  MATHEMATK  s 


450.  Theorem  of  Pythagoras  proved.  No  doubt  the 
student  now  sees  that  the  theorem  of  Pythagoras  is 
proved  by  Exs.  1~5,  aboye.  We 
shall,  however,  set  up  the  prooi' 
for  reference.  l>, 


(liven  the  right  triangle  A  /'><'. 
right-angled  at  C,  to  prove  that  the 
square  on  the  hypotenuse  is  equal 
to  the  sum  of  the  squares  on  the 
sides  including  the  right  angle.  In  terms  of  Fig.  292  this 
means  to  prove  that  c2  =  a2  -\- l>~. 


Proof 


STATEMENTS 


REASONS 


In    Fig.  292    draw   CI)±AJi 
and  letter  the  figure  as  shown. 


rr, 

Then 


and 


(1) 

(2) 


In   (1)   and   (2)   I-  =  me  and 
=  nc.  (3) 


By  adding  the  two  equations 
in  (3), 

a-  +  b-  =  me  +  ne.     (4) 
a2  +  ft2=±e(jn  +  n).  (5) 

But        m  +  »  —  c.  ((5) 


.-.  n-  +  //-  =  ,--. 


(7) 


Because  if  in  a  right  triangle 
a  line  is  drawn  from  the  vertex 
of  a  right  angle  perpendicular  to 
the  hypotenuse,  either  side  about 
the  right  angle  is  a  mean  pro- 
portional between  the  whole  hy- 
potenuse and  the  segment  of  the 
hypotenuse  adjacent  to  it. 

Because  when  four  quantities 
are  in  proportion  the  product  of 
the  me'ans  equals  the  product  of 
the  extremes. 


Addition  axiom. 
By  factoring  out  c. 
The  whole  is  equal  to  the  sum 
of  all  its  parts. 
Bv  substitution. 


PYTHAGORAS 


402  GENERAL  MATHEMATICS 

HISTORICAL  NOTE.  Pythagoras  (c.  569  u.c.-c.  500  B.C.),  the  second 
of  the  great  philosophers  of  Greece,  is  said  to  hate  "  changed  the  study 
of  geometry  into  the  form  of  a  liberal  education."  After  some  wander- 
ings, he  founded  the  famous  Pythagorean  School  at  Croton,  a  Dorian 
colony  in  the  south  of  Italy.  Here  enthusiastic  audiences  composed 
of  citizens  of  all  ranks,  especially  the  upper  classes,  crowded  to  hear 
him.  It  is  said  that  the  women  went  to  hear  him  in  direct  violation 
of  a  law  against  their  public  appearance. 

'  Pythagoras  divided  his  audiences  into  two  classes :  the  Proba- 
tioners (or  listeners)  and  the  Pythagoreans  (or  mathematicians). 
After  three  years  in  the  first  class  a  listener  could  be  initiated 
into  the  second  class,  to  whom  were  confided  the  main  discoveries 
of  the  school. 

The  Pythagoreans  formed  a  brotherhood  in  which  each  member 
was  bound  by  oath  not  to  reveal  the  teachings  or  secrets  of  the 
school.  Their  food  was  simple,  their  discipline  severe,  and  their 
mode  of  life  arranged  to  encourage  self-command,  temperance,  purity, 
and  obedience. 

The  triple  interwoven  triangle,  or  pentagram  (star-shaped  regular 
pentagon),  was  used  as  a  sign  of  recognition,  and  was  to  them  a  symbol 
of  health.  It  is  related  that  a  Pythagorean  while  traveling  fell  ill  and, 
although  carefully  nursed  by  a  kind-hearted  innkeeper,  was  unable 
to  survive.  Before  dying,  however,  he  inscribed  the  pentagram  star 
on  a  board  and  begged  his  host  to  hang  it  up  outside.  This  the  host 
did ;  and  after  a  considerable  length  of  time  another  Pythagorean, 
passing  by,  noticed  the  sign  and,  after  hearing  the  innkeeper's  story, 
rewarded  him  handsomely.  One  motto  of  the  brotherhood  was : 
"A  figure  and  a  step  forwards;  not  a  figure  to  gain  three  oboli." 

The  views  of  society  advocated  by  the  brotherhood  were  opposite 
to  those  of  the  democratic  party  of  Pythagoras's  time,  and  hence 
most  of  the  brotherhood  were  aristocrats.  For  a  short  time  the 
Pythagoreans  succeeded  in  dominating  affairs,  but  a  popular  revolt 
in  501  B.C  led  to  the  murder  of  many  prominent  members  of  the 
school,  and  Pythagoras  himself  was  killed  shortly  afterwards. 

Though  the  brotherhood  no  longer  existed  as  a  political  party,  the 
Pythagoreans  continued  to  exist  a  long  time  as  a  philosophical  and 
mathematical  society,  but  to  the  end  remained  a  secret  organization, 
publishing  nothing,  and  thus  leaving  us  little  information  as'to  the 
details  of  their  history.  See  Ball's  "A  History  of  Mathematics,"  p.  19. 


EXERCISES 

1.  The  base  and  altitude  of  a  right  triangle  (Fig.  293)  are 
6  and  8  respectively.    What  is  the  length  of  the  hypotenuse  ? 
(Use  the  theorem  of  Pythagoras.) 

2.  How  long  must  a  rope  be  run  from  the 
top  of  a  16-foot  tent  pole  to  a  point  20  ft.  from 
the  foot  of  the  pole  ? 

3.  A  baseball  diamond  is  a  square  a  side  of 

which  is  90  ft.    What  is  the  length  of  a  throw 

_b  IG.  —  t'f> 

from  "  home  "  to  "  second  "  ? 

4.  Find  the  formula  for  the  diagonal  of  a  square  whose 
side  is  s.    Use  this  formula  to  determine  the  diagonal  when 
s  =  10 ;  when  s  =  15. 

5.  Prove  from  the  Pythagorean  theorem  that  a  =  Vc2  —  62 
and  translate  the  equation  into  words. 

6.  Prove  also  that  b  =  Vc2  —  a2  and  translate  the  equation 
into  words. 

7.  A  ladder  20ft.  long  just  reaches  a  window  15ft.  above 
the  ground.    How  far  is  the  foot  of  the  ladder  from  the  foot 
of  the  wall  if  the  ground  is  level  ? 

8.  The  hypotenuse  of  a  right  triangle  is  35  ft.  and  the 
altitude  is  21  ft.    Find  the  base. 

9.  Using  the  formula  of  Ex.  6,  find  the  value  of  a  when 
c  =  22  and  b  =  20. 

10.  A  tree  standing  on  level  ground  was  broken  24  ft.  from 
the  ground,  and  the  top  struck  the  ground  18  ft.  from  the  stump, 
the  broken  end  remaining  on  the  stump.    How  tall  was  the  tree 
before  breaking  ? 

11.  Construct   on    squared    paper   a   right   triangle,    using 
the  following   pairs    of   numbers    for  the  base  and  altitude 


404 


GENERAL  MATHEMATICS 


respectively  :  1  and  1 ;  1  and  V2 ;  2  and  2 ;  2  and  3 ;  4  and  4 ; 
4  and  5 ;  1  and  5 ;  2  and  5  ;  3  and  5  ;  12  and  1. 

HINT.    Use  the  line  segment  you  obtained  for  the  first  part. 

12.  Calculate  for  each  part  of  Ex.  11  the  length  of  the 
hypotenuse. 

451.  The  theorem  of  Pythagoras  furnishes  a  method  of 
constructing  with  ruler  and  compasses  the  square  root  of  a 
number.  Exercises  11  and  12,  Art.  450,  suggest  a  method 
of  finding  the  square  root  of  a  number  by 
means  of  ruler  and  compasses.  The  method 
is  illustrated  by  the  following  exercise : 

Construct  the  square  root  of  42. 

The  following  study  (analysis)  of  the  exer- 
cise will  help  us  to  understand  the  problem. 

Suppose  that  we  have  the  figure  constructed ; 
that  is,  let  us  imagine  that  Fig.  294  is  the  required 
figure  and  that  AB  is  the  required  length  V42. 

Now  a  and  b  can  be  of  various  lengths  provided 
a-  +  tf  =  42.  (Why?)   Let  us  suppose  that  CB  -  6  ; 
then  how  long  is  l>'l    We  know  that  o(>  +  tr  —  42 
would  be  the  equation  from  which  the  value  of  b  can  be  found. 
It  is  clear  that  b  would  have  to  equal  A7/6. 

Then  the  problem  merely  becomes  one  of  learning  how  to  con- 
struct V6.  Some  members  of  the  class  may  already  know  how 
to  do  this,  but  we  shall  proceed  with  our  analysis. 

"Imagine  another  triangle,  A'B'C'  (Fig.  295),  so  con- 
structed that  the  hypotenuse  turns  out  to  be  V6  and  so 
that  B'C'  is  2  units  long;  then  A  'C'  must  equal  V2.  Why? 

Our  problem  finally  reduces,  then,  to  a  problem  of 
constructing  V2.  If  we  can  find  this  geometrically  we 
can  solve  the  original  exercise,  as  our  analysis  has  shown. 

We  already  know  how  to  construct  v2  by  constructing  a  right 
triangle  with  the  two  legs  about  the  right  angle  equal  to  1.  Then 
the  hypotenuse  equals  V2.  Why? 


INTERPRETATION  OF  ROOTS  AND  POWERS    405 

We  then  reverse  our  analysis  as  follows : 

(a)  Construct  V2  as  indicated  above. 

(b)  Construct  a  second  right  triangle  with  a  base  of  v'2  units  long 
and  an  altitude  2  units  long.    Its  hypotenuse  will  equal  Vti.    Why? 

(c)  Construct  a  third  right  triangle  whose  base  is  \/6  units  long 
and  whose  altitude  is  6  units  long.  Its  hypotenuse  will  be  \/42  units. 

Why? 

EXERCISE 

Construct  with  compasses  a  line  segment  equivalent  to  each  of 
the  following:  V6  ;  VTT  ;  A/27  ;  Vl43;  V214  ;  3  VJJ ;  2  V2. 

452.  Mean  proportional  construction  a  method  for  find- 
ing square  roots.  We  shall  now  see  that  our  mean  pro- 
portional construction  (Art.  374)  furnishes  us  with  an  easy 
method  of  constructing  square  roots. 

EXERCISES 

1.  Review  the  construction  (Art.  374)  for  finding  a  mean 
proportional  between  two  line  segments  a  and  b  (Fig.  296). 

2.  Construct  the  mean  proportional          a 

between  4  and  9 ;  4  and  16.  ft 


3.  Review  the  proof  for  the  statement 

..        , T    ,  FJG.  296 

that  a  mean  proportional  between  two  line 

segments  a  and  b  equals  the  square  root  of  the  product  of  a  and  //. 

The  preceding  exercises  suggest  that  the  mean  propor- 
tional construction  furnishes  a  method  for  finding  the  square 
root  of  a  number.  For  example, 

Find  the  square  root  of  12. 

On  squared  paper  find  the  mean  propor-  . 
tional  of  two  factors  of  12,  for  example,  2  / 
and  6.  The  mean  proportional  x  (Fig.  297)  {_ 
is  the  square  root  of  12,  for  K—  2- 

1  —  -.  Why?  FIG.  297.   MEAN  PROPOR- 

y      6  TIONAL  METHOD  OF  FIND- 

x2  =  12^_  Why?  ING  THE  SQUARE  ROOT  OF 

Whence            x  =  Vl2.  Whv?  A  NUMBER 


406  GENERAL  MATHEMATICS 

EXERCISE 

Construct  the  square  root  of  21 ;  6  ;  5  ;  18  ;  42 ;  84  ;  66  ; 
76.  Compare  the  results  with  the  table  of  Art.  449.  Your 
results  ought  to  approximate  the  second  decimal  place. 

453.  Large  numbers  under  the  radical  signs.  When  the 
number  under  the  radical  sign  is  large,  the  various  geo- 
metric constructions  for  finding  square  roots  are  neither 
convenient  nor,  in  general,  sufficiently  accurate.  In  this 
case  it  is  of  advantage  to  reduce  the  given  quadratic  surd 
to  an  equivalent  expression  which  has  a  smaller  number 
under  the  radical  sign.  Suppose  we  wish  to  find  the  value 
of  V5056.  The  square  root  is  at  once  evident  if  we  resolve 
the  number  into  two  equal  groups  of  factors ;  thus : 


V5056  =  V(22  •  3  .  7)  (22  •  3  .  7)  =  V84  •  84  =  84. 

Even  when  the  number  is  not  a  perfect  square,  factor- 
ing will  often  enable  the  student  to-  find  its  square  root 
much  more  easily,  as  will  be  shown  later. 

EXERCISES 

Find  the  following  indicated  square  roots  : 

1.  V576.  3.  V484.  5.  V3600. 

2.  V1296.  4.  V1089.  6.  Vl936. 

454.  The  square  root  of  a  product.  The  preceding  exer- 
cises show  that  the  square  root  of  the  product  of  several 
factors,  each  of  which  is  a  square,  may  be  found  by  taking 
the  square  root  of  each  factor  separately,  as  in  the  follow- 
ing examples: 

1.  V9-  25=V9V25  =  3-  5=15. 


INTERPRETATION  OF  ROOTS  AND  POWERS    407 

This  is  true  because  9  •  25  can  be  written  as  the  product 
of  two  groups  of  equal  factors  (3  •  5)  (3  •  5).  Hence,  by  the 
definition  of  a  square  root,  (3  •  5)  is  the  square  root  of  9  •  25. 

2.   Vl6  xY  =  Vl6  V^  Vp  =  4  a*f. 

This  is  true  because  16  afy6  may  be  written  as  the  product 
of  two  groups  of  equal  factors  (4  a^?/3)  (4  z2?/3).  Hence 
4  a;2?/3  is  the  square  root  of  IGa^y6. 

The  preceding  exercises  show  that  the  square  root  of  a 
product  is  obtained  by  finding  the  square  root  of  each  factor 
separately  and  then  taking  the  product  of  these  roots.  That 

is,  in  general,  _  _ 

v  a  •  b  =  v  a  •  Vft. 

This  principle  may  be  used  to  simplify  radical  surds  in 
the  following  manner.  Suppose  we  wish  to  find  the  value 

of  V11858.    Then  _ 

V11858  =  V2  •  11-  11  •  7-7 


=  77  V2. 

By  the  table  of  roots,  V%  =  1.414. 

Then  77(1.414)  =  108.878. 

Hence  V11858  =  108.878. 

It  will  be  helpful  to  observe  the  following: 

(1)  The  principle  enables  us  to  simplify  the  radicand  to 
a  point  where  we  can  easily  find  the  root  by  the  table  or 
by  several  geometric  constructions. 

(2)  A  quadratic  surd  is  in  its  simplest  form  when  the 
number  under  the  radical  sign  does  not  contain  a  perfect 
square  factor. 

In  general,  if  the  expression  under  the  radical  sign 
contains  a  factor  which  is  a  square,  this  factor  may  be 
removed  by  writing  its  square  root  before  the  radical  sign. 


-108  GENERAL  MATHEMATICS 

EXERCISES 

Change  the  following  so  as  to  leave  no  factor  which  is  a 
square  under  the  radical  sign : 


1.   Vl2. 

_6.   V2S. 

11. 

2.   V40. 

7.   V75. 

12. 

3.   VlS. 

8.   V125. 

v  13. 

4.   V50. 

^9.   VlOS. 

-  14. 

'   5.   V72. 

10.   Vo*. 

*  15. 

455.  Value  of  memorizing  square  roots  of  certain  numbers. 
Exercises  1~9,  Art.  454,  suggest  the  manner  in  which  the 
square  roots  of  a  few  small  numbers,  like  2,    3,  5,  are 
made  to  do   service  in  finding  the  roots   of  many  large 
numbers.    In  fact,  many  students  in  other  fields  find  that 
memorizing  these  numbers,  which  occur  again  and  again 
in  their  problems,  increases  their  efficiency. 

EXERCISES 

From  the  table  of  roots  we  know  that  V2=1.414,  A/3=1.732, 
and  V 5  =  2.236.  Using  these  facts,  compute  each  of  the  fol- 
lowing correct  to  two  decimal  places  : 

1.  V75.  4.   V72.  7.   V50  +  V75-V6. 

2.  V80.  5.   V98.  8.  2  V32  +  V72  -  Vl8. 

3.  V48.  6.   V363.  9.   V45-\/|. 

D 

456.  The  square  root  of  a  fraction.   A  fraction  is  squared 
by  squaring  its  numerator  and  its  denominator  separately 

9 

and  indicating  the  product  thus :        x  -  =  —.    Hence,  to 

bob 

extract  the  square  root  of  a  fraction,  we  find  the  square 
root  of  its  numerator  and  denominator  separately. 
For  example,  VT\  =  J,  since  %  •  %  =  f  V- 


INTERPRETATION  OF  ROOTS  AND  POWERS    409 

EXERCISES 

Find  the  square  roots  of  the  following  accurate  to  two  places 
(use  the  table  for  square  roots): 

1.  f.     2.  f.     3.  TV     4.  f.     5.   f.     6.  V-.      7.  _*_.     8.   _2y. 

457.  Rationalizing  the  denominator  of  a  fraction.  It 
probably  has  been  noticed  that  the  calculation  of  those 
problems  of  the  preceding  lists  in  which  the  denominator 
is  a  square  is  far  easier  than  when  this  is  not  the  case. 
This  may  be  illustrated  as  follows: 


[2  _  V2 


1.414 
1.732 


Dividing  1.414  by  1.732  is  not  an  easy  division.   In  fact, 
most  people  would  find  it  impossible  to  do  mentally.    On 

the  other  hand,  J?  =  _£  =  ?dii?  =  Q.816  involves  an  easy 

M9      V9          8 
mental  division. 

It  is  of  interest  to  note  that  the  denominator  of  a 
fraction  can  always  be  made  a  square  merely  by  multi- 
plying numerator  and  denominator  by  the  proper  factor  ; 


,   Jf  =  J?  =  ^  = 
i  o         \  9         o 


thus,  =          =        =  =  0.816.     Note    that   the 

o 

division  -—  —  is  an  easy  mental  division  compared  with 
o 

the  division    '        ,  which  we  had  above. 


A\re  may  summarize  this  method  of  taking  the  square 
root  of  a  fraction  as  follows  : 

1.  To  find  the  square  root  of  a  fraction,  change  the  fraction 
into  an  equivalent  fraction  whose  denominator  is  a  perfect 
square. 


410  GENERAL  MATHEMATICS 

2.  The  square  root  of  the  netv  fraction  equals  the  square 
root  of  its   numerator   divided    by    the    square    root    of  its 
denominator. 

3.  If  desired,  express  the  result  in  simplest  decimal  form. 

The  process  of  changing  a  radical  expression  so  as  to 
leave  no  denominator  under  a  radical  sign  is  called  ration- 

alizing the  denominator.    Thus,  -yj  -  =  -  V3  ;  \J-  =  -^a- 

1  3      o  \  a      a 

EXERCISES 

1.  Find  the  value  of  the  following  square  roots  (find  values 
approximately  accurate  to  the  second  decimal  place)  : 

(a)  J.      (b)  f  .      (c)  1       (d)  1       (e)  f  .      (f)  f      (g)  TV 

2.  Rationalize  denominators  of  the  following,  and  simplify  : 


3.  What  is  the  value  of  the  expressions  in  Ex.  2  from  (e) 
to  (j),  inclusive,  if  a  =  3  and  b  =  2  ? 

458.  Addition  and  subtraction  of  surds.  Sometimes  the 
arithmetic  in  a  problem  may  be  simplified  if  the  surds  are 
combined  into  one  term.  Thus, 

V20  +  V45  =  2  V5  +  3  V5 

=  5V5,  or5(2.236) 
=  11.180. 

By  adding  2  \/5  +  3  V5  just  as  we  add  2_-  4  plus  3  •  4,  we 
need  only  to  look  up  the  \/5,  whereas  V20  +  V45  calls  for 
two  square  roots. 


INTERPRETATION  OF  ROOTS  AND  POWERS    411 

In  the  following  list  simplify  each  expression  as  far  as 
possible  without  using  approximate  roots  ;  that  is,  leave 
your  result  in  indicated  form.  Practically  this  is  often 
better  than  finding  an  approximation,  for  in  this  manner 
you  submit  results  that  are  absolutely  accurate.  It  leaves 
the  approximation  to  the  next  person,  who  may  find  as 
many  decimal  places  as  the  needs  of  his  particular  problem 
demand. 

EXERCISES 

1.  Vl08  +  V75  +  Vl2.  8.   » 

O 

2.  2V98-VT8.  9.  Vf  + 

3.  6  V288  -  4  Vl8  +  Vl28.     10.  Vf- 

4.  5  V432  -  4  V3  +  Vl47.       n.  1+^ 

3 

5.  2  V27  +  3  V48  -*3  V75.     12.  -  f  + 

6.  3  V20  +  2  Vl25  -  Vl80.     13.  ^/T  _  2  V2  + 

4-  V-  14.  4  V28  +  3  Vo5  -  Vll2. 


459.  Multiplication  and  division  of  quadratic  surds.  These 
two  processes  will  be  treated  briefly.  In  elementary  mathe- 
matics there  are  few  verbal  problems  which  involve  this 
process.  Further,  the  principles  involved  do  not  offer  any- 
thing new  for  us. 

Thus,  to  divide  V2  by  V5,  we  may  write  this  in  the 

form  of  a  fraction  —  —  and  proceed  as  we  ordinarily  do 

when    finding    the    value    of    a    fraction    which    involves 
quadratic   surds. 

The  rule  in  multiplication  is  equally  familiar.  The 
equation  Vab=\/aVb  may  be  read  just  as  well  from 
right  to  left,  and  we  have  VaV5=VaJ.  Thus,  V2V5  is 
precisely  the  same  as  VlO. 


n-2  GENERAL  MATHEMATICS 

EXERCISES 

1.  Find  the  product  of  the  following: 

(a)  V3  V27.  (d)  V(5Wl2a;8~. 

(b)  vW*8.  (e)  V|VH-_ 

(c)  V3V5.  (f)  V|Vf  Vf  Vl 

(g)  ( V2  +  Va  -  Vs)(  V2  -  Vs  +  Vs). 

Solution.  We  multiply  each  term  of  one  polynomial  by  each  term 
of  the  other  and  simplify  the  result  as  follows : 

V2  +  V§  -  V5 
VQ  -  Va  +  Vs 
2  +  Ve  -  Vio 
-  Ve          -  a  +  Vis 
+  VTo       +  Vis  -  5 


2  Vlo  -  6 

(h)  (2 Vr  +  Vs  -  Vs)(  Vs). 

(i)  (4V2-2V3  + V5)(2  V3). 

(j)  (Vs  -  3 V2  +  VTo)(Vs). 

(k)  (3  V2  +  4  Vs)(2  V2  -  S  Vs). 

(i)  ( Vs  -  VI  +  Vs)(V3  +  VI  -  Vs). 

2.  Divide,  and  express  the  result  in  simplest  form : 

(a)  1  by  VS.  (c)  V6  by  Vs. 

(b)  24  by  Vs.  (d)  2  Vl2  -  5  VlS  by  Vs. 

460.  Fractional  exponents  another  -means  of  indicating 
roots  and  powers.  Thus  far  we  have  not  used  a  fraction 
as  an  exponent,  and  evidently  it  could  not  be  so  used 
without  extending  the  meaning  of  the  word  "exponent." 

Thus,  a?  means  x  •  x  •  x,  but  Xs  evidently  cannot  mean 
that  x  is  to  be  used  as  a  factor  one  half  of  a  time. 


INTERPRETATION  OF  ROOTS  AND  POWERS    413 

It  is  very  important  that  all  exponents  should  be  gov- 
erned by  the  same  laws;  therefore,  instead  of  giving  a 
formal  definition  of  fractional  exponents,  we  shall  lay  down 
the  one  condition  that  the  laws  for  integral  exponents  shall 
be  generally  true,  and  we  shall  permit  the  fractional  ex- 
ponent to  assume  a  meaning  necessary  in  order  that  the 
exponent  laws  shall  hold. 

Since  we  agree  that  ^  •  x^  —  x,  we  see  that  a?*  is  one  of 
the  two  equal  factors  of  x ;  that  is,  x^  is  the  square  root 
of  #,  or  we  may  write  y?  =  V#. 

Similarly,  since  Xs  •  Xs  •  x5  =  z,  Xs  is  the  cube  root  of  x ; 

\  3/ — 

that  is,  Xs  =  v  x. 

Again,  x*  •  x*  •  x'5  =  Xs  =  xt.    This  means  that  Xs  is  the 

2  3/ 

cube  root  of  «2 ;  that  is,  x*  =  v  x2. 

This  discussion  is  sufficient  to  show  that  the  fractional 
exponent  under  the  laws  which  govern  integral  exponents 
takes  on  a  meaning  which  makes  a  fractional  exponent 
just  another  way  of  indicating  roots  and  powers ;  that  is, 
the  denominator  indicates  the  root,  and  the  numerator  indi- 
cates the  power. 

Thus,  8^  means  to  take  the  cube  root  of  8  and  square 
the  result,  or  square  8  and  take  the  cube  root  of  the  result. 
In  either  case  the  final,  result  is  4.  Again,  10*  (or  10°-6666) 
means  to  take  the  cube  root  of  100,  which,  by  the  table  of 
Art.  449,  is  4.641.  Note  that  a  common-fraction  exponent 
may  appear  as  a  decimal  fraction. 

From  this  it  will  be  obvious  that  the  student  needs  only 
to  become  familiar  with  the  new  method  of  writing  roots 
and  powers.  The  following  brief  list  of  problems  presents 
precisely  the  same  ideas  as  those  in  the  preceding  list 
dealing  with  surds ;  only  the  form  is  different. 


414  GENERAL  MATHEMATICS 

EXERCISES 
1.  Write  in  simplest  form  : 

(a)  4*  +  9*  +  16*  +  25*  +  36*. 

(b)  1*  +  8*  +  64*  +  0*. 

(c)  64*  +  9*  +  16*  +  (-  32)*  -  (-  27)1 

(d)  24*  +  54'  -  6*. 

(e)  18*  +  32*  -  Vl28  +  V2. 

(f)  (^*+8*-(f)4+(50)*. 

(g)  (81)*  -  2  (24)*  +  V28  +  2  (63)*. 


2.  Multiply: 

(a)  z*z*.  (c)  102-5  •  10°-.125  •  101-25. 

(b)  10*  •  10*  -  10*.  (d)  103-6250  •  10°-3750  .  10°-0625. 

3.  Translate  (c)  and  (d),  Ex.  2,  into  expressions  with  common 
fractions  as  exponents  and  estimate  how  large  the  numbers 
would  be  if  we  had  a  way  of  finding  the  result. 

NOTE.  100-8"5  means  10^,  or  the  eighth  root  of  1000.  Another 
way  to  look  at  it  is  that  10  is  to  be  raised  to  the  375th  power  and 
the  thousandth  root  taken.  These  ideas  are  of  very  great  importance 
in  getting  a  clear  understanding  of  the  chapters  on  logarithms  and 
the  slide  rule.  In  these  chapters  we  shall  learn  how  to  find  a  root 
(say  the  15th)  just  as  easily  as  the  square  root. 

461.  Zero    exponents.    Under    the    laws    which   govern 
integral  exponents  a°  =  1,  as  is  shown  by  the  following  : 
«5-=-  a5=  a°.   (By  the  Division  Law.) 
o5-5-o6  =  l.    (The  quotient  of  any  number  divided 

by  itself  is  1.) 
Hence     a°  =  l.    (By  the  equality  axiom.) 

Thus  the  zero  power  of  any  number  (except  zero)  is  1. 
Thus,  15°  =1;  (560)°=  1;  (-6a:)0=l;  10°  =  1. 


462.  Negative  exponents.    If  x~s  obeys  the  same  law  as 
integral  exponents,  then  • 

_  Q  Q  Q  -| 

-3-  X  —  =  —  ,     or     —  ; 

1  X6        X6  X6 

that  is,  x~  s  —  — 

Similarly,  if  we  multiply  the  numerator  and  denominator 

x~a  1 

of  -    -  by  xa,  we  obtain  x~a  =  —  • 

Tin-,  ~~>  »  looAo-^jL,  v-'-mf 

J^  =  10,000. 

See  if  you  can  state  in  simple  language  the  meaning  of 
a  negative  exponent. 

EXERCISES 

Simplify : 

1.  10- 2  x  103-125  x  100-0625  x  10- 1-03675. 

2.  10^  x  lo"  x  (56)°  x  10*. 

3.  (39)  (169)"*.  6.  16- 2-4.  9.  xm  •  xn. 

4.  1000  •  (100)"*  7.  1000- IO-3.  «      « 

i  in     yb  .  yc 

5.  2X  x  2-*  x  2.  8.  144~*  •  24. 

11.  XmH-Xn. 

12.  (+625)*;  (-125)-*. 

13.  (x2)3;    (x3)4;   (x0)2. 
HINT,    (z2)3  =  x2  •  xz  •  x2. 

14.  (IO2)2;  (IO2)8;  (IO2)4. 

15.  (10°-125)2;    (100-0625)4;   (10°-125)8. 

16.  (xm)n ;  translate  the  formula  (xm}n  =  xmn  into  an  alge- 
braic rule. 

17.  -v/x^;    ^x8;    ^x5. 

NOTE.    The  3  in  vx®  means  "find  the  cube  root." 


416 


GENERAL  MATHEMATICS 


18. 

19.  Since  (xm)n  ^  xmn,  what  is 

m 

Translate  the  formula  Vic™  =  x"  into  an  algebraic  rule. 

20.  Find  the  value  of : 

(a)  VlO^.      (b)  A/10"-'75,      (c)  -v/H)3-"250.      (d)  -x/lO5-8750. 

463.  Cube  of  a  binomial.  We  shall  now  see  how  certain 
other  short  cuts  in  finding  powers  may  be  illustrated  and 
explained. 

INTRODUCTORY  EXERCISES 

1.  Find  the  cube  of  x  -f-  y  by  first  finding  the  square  of 

x  -f  y  and  multiplying  the  result  by  x  +  y. 

2.  By  multiplication  find  (a  +  &)3. 

3.  Find  the  cube  of  (x  —  y);    of  (a  —  &).    Compare  these 
cubes    with    the    results    of    Exs.  1 

and  2. 

4.  It  will  be  helpful  to  your  class- 
mates if  you  will  make  a  set  of  blocks, 
as   shown   in   Fig.  298,    in    order   to 
show  that 

(x  -f-  7/)3  =  x3  +  3  x2y  +  3  xif  +  ?/3- 
How  many  blocks  are  needed  ? 

5.  Find  the  following  cubes,  doing 
as  many  as  you  can  mentally : 

(a)  (c  +  dy.    (c)  (c-df.      (e)  (a-4-27/)8.   (g)  (2x  + 

(b)  (m  +  nf.    (d)  (m-nY-    (f)  (2 x  -  y)s.  (h)  (2x-3y)s. 

Exercises  1—5  show  that  the  cube  of  a  binomial  is  equal 
to  the  cube  of  the  first  term,  plus  three  times  the  square  of 
the  first  term  multiplied  by  the  second,  plus  three  times  the 
first  term  multiplied  by  the  square  of  the  second,  plus  the 
cube  of  the  second  term. 


SIR  ISAAC  NEWTON 


418  GENERAL  MATHEMATICS 

HISTORICAL  NOTE.  The  first  use  of  positive  and  negative  frac- 
tions as  exponents  is  found  in  a  book  written  by  the  great  English 
mathematician  and  physicist,  Sir  Isaac  Newton  (1642-1727).  He 
discovered  the  binomial  theorem  and  numerous  laws  of  physics ; 
for  example,  the  law  of  gravitation,  the  law  about  lenses  and  prisms, 
and  the  explanation  of  the  rainbow.  Among  his  numerous  books  are 
"  Universal  Arithmetic  "  (really  an  algebra)  and  "  Principia  "  (one  of 
the  greatest  books  of  all  times). 

The  biography  of  Xewton  (see  Ball's  "  A  Short  History  of  Mathe- 
matics," pp.  328-362,  and  Cajori's  "A  History  of  Elementary  Mathe- 
matics," pp.  238-240)  is  very  interesting  and  inspiring.  As  a  boy 
he  was  expected  to  be  learning  how  to  take  care  of  his  father's 
farm,  but  he  spent  much  of  his  time  studying  and  trying  mechan- 
ical experiments.  Thus,  we  read  of  his  constructing  a-  clock  run 
by  water  which  kept  very  fair  time.  His  mother  noticing  this 
sensibly  resolved  to  send  him  to  Cambridge.  Here  followed  a  brilliant 
career  of  thirty-five  years  as  student  and  teacher.  As  a  professor  it 
was  his  practice  to  lecture  publicly  once  a  week,  and  then  only  from 
half  an  hour  to  an  hour  at  a  time.  In  the  week  following  he  gave 
four  hours  of  consultation  to  students  who  wished  to  discuss  the 
results  of  the  previous  lecture.  It  is  said  that  he  never  repeated  a 
course  and  that  one  course  began  at  the  point  where  the  preceding 
course  had  ended.  The  result  of  his  second  study  during  this  period 
has  dazed  master  minds  who  have  attempted  to  understand  all  that 
Newton  accomplished.  Perhaps  you  will  later  agree  with  some  of 
the  following  tributes  to  him  : 

Nature  and  Nature's  laws  lay  bid  in  night  : 

God  said,  "  Let  Newton  be  !  "  and  all  was  light.  —  POPE 

There,  Priest  of  Nature  !  dost  thou  shine, 

Newton !  a  king  among  the  kings  divine.  —  SOCTHEY 

Tafcing  mathematics  from  the  beginning  of  the  world  to  the  time  when 
Newton  lived,  what  he  had  done  was  much  the  better  half.  —  LEIBNITZ 

I  don't  know  what  I  may  seem  to  the  world,  but  as  to  myself,  I 
seem  to  have  been  only  as  a  boy  playing  on  the  sea-shore,  and  diverting 
myself  in  now  and  then  finding  a  smoother  pebble  or  a  prettier  shell 
than  ordinary,  whilst  the  great  ocean  of  truth  lay  all  undiscovered 
before  me. —  NEWTON 


INTERPRETATION  OF  ROOTS  AND  POWERS    419 


464.  Cube  roots.  The  equation  y  =  $  asserts  that  y 
is  the  cube  of  .r,  or,  inversely,  x  is  the  cube  root  of  y. 
If  the  equa- 
tion is  graphed 
we  shall  ob- 
tain a  curve 
analogous  to 
the  curve  for 
squares  and 
square  roots 
which  may  be 
used  to  find 
cube  roots  and 
cubes.  We  pro- 
ceed to  find 
corresponding 
values  for  y 
and  x  in  order 
that  we  may 
plot  sufficient 
points  for  the 


-x- 


-y- 


curve. 


FIG.  299.    GRAPH  OF  y  -  Xs 


EXERCISES 

1.  What  is  the  value  of  y  in  the  equation  y  •=  y?  when  x 
equals  0  ?    when  x  =  + 1  ?   when  x  —  —  1  ?    when   x  =  +  2  ? 
when  x  =  —  2  ? 

2.  Calculate  values  as  in  Ex.  1  and  fill  the  blank  spaces  of 
the  following  table.    If  the  curve  is  not  obvious,  expand  the 
table  until  you  have  enough  points  to  draw  the  curve. 


X 

y 

0 

+  1 
+  1 

+  2 
+  8 

-2 
-8 

±3 

±4 

±6 

±6 

±7 

±i 

±i 

±§ 

±f 

±1 

±&. 

±1. 

0 

±27 

420  GENERAL  MATHEMATICS 

From  this  table  we  may  obtain  the  curve  in  Fig.  299. 
One  small  square  vertically  represents  1  unit,  and  5  small 
squares  horizontally  represent  1  unit. 

From  this  curve  we  can  read  off,  approximately,  the 
cube  or  the  cube  root  of  any  number;  thus  the  cube  of 
2.2  is  seen  to  be  about  10.5  (by  the  table  actually  10.64); 
the  cube  root  of  13  is  seen  to  be  a  little  over  2.4  (see 
the  table  for  values  accurate  to  two  decimal  places).  More 
accurate  results  can  be  obtained  by  drawing  the  curve  to 
a  large  scale. 

465.  Cube  roots  of  arithmetical  numbers.   An  arithmetical 
method  for  finding  cube  roots  based  on  an  algebraic  for- 
mula for  (a  +  6)3  could  be  devised.    But  this  method  is 
seldom  used,  because  cube  roots  as  well  as  higher  roots 
are   more   quickly  found   by  means   of  logarithms.    This 
method  will  be  taught  in  the  next  chapter.    In  the  mean- 
time the   student  may  for  all  practical  purposes  use  the 
table  in  Art.  449. 

Furthermore,  we  could  devise  analogous  rules  and  curves 
for  fourth  roots,  fifth  roots,  fourth  powers,  and  so  on,  but 
these  too  are  more  readily  found  by  logarithms. 

466.  Indicating  higher  roots.    By  means  of  an  index  figure 
the  radical  sign  is  made  to  indicate  other  roots  than  square 
roots. 

Thus  the  cube  root  of  8  or  one  of  its  three  equal  factors 
is  written  V8  =  2.  The  fourth  root  of  16  is  written  Vl6  =  2. 
The  3  in  v  8  is  the  index  of  the  root. 

Any  expression  which  contains  an  indicated  root  is  called 
a  radical  expression. 

The  principles  of  reducing  surds  to  simpler  forms,  dis- 
cussed in  detail  for  quadratic  surds,  may  be  applied  to 
higher  indicated  roots. 


EXERCISES 

1.  Simplify  the  following  (remove  any  factor  which  is  a 
perfect  power  of  the  degree  indicated  by  the  index): 

^32;  -\/64;   ^64  x6/;  "\/48  «4  ;   A/16  a 

2.  Add  and  subtract  as  indicated: 

(a)  A/16  +  -^54  -  -^250  +  -v 

(b)  -v/54  +  -v/128  +  -v/1024  + 


SUMMARY 

467.  This  chapter  lias  taught  the  meaning  of  the  fol- 
lowing words    and    phrases  :    square    root   of    a   number, 
quadratic  surd,  radical  sign,  radicand,  quadratic  trinomial 
square,  index. 

468.  The  graph  of  the  equation  y  =  x2  was  used  as  a 
device  for  finding  squares  and  square  roots. 

469.  A  positive  number  has  two  square  roots  ;    thus, 
V4  =  +  2  or  -  2. 

470.  The  square  of  the  sum  or  difference  of  two  numbers 
may  be  found  by  the  formula 

(a  ±  A)2  =  a2  ±  2  db  +  ft2. 
This  formula  was  illustrated  geometrically. 

471.  To  find  the  square  root  of  a  perfect  square  tri- 
nomial: Extract  the  square  roots  of  the  two  perfect  square 
terms  and  connect  them  by  the  sign  of  the  remaining  term. 

472.  The  square  of  a  trinomial  consists  of  the  sum  of  the 
squares  of  its  terms  plus  twice  the  product  of  each  term  by 
each  succeeding  term.    By  remembering  this  rule  the  square 
roots  of  some  polynomials  may  be  found  by  inspection. 


422  GENERAL  MATHEMATICS 

473.  The  chapter  taught  a  method  of  finding  the  square 
root  of  algebraic  polynomials  and  arithmetical  numbers. 

474.  The   chapter   includes    a    simple    table    of    square 
roots  and  cube  roots. 

475.  Quadratic  surds  may  often  be  simplified  by  apply- 
ing the  principle  vW>=VaVi. 

476.  We  find  the  square  root  of  a  fraction  by  dividing 
the  square  root  of  the  numerator  by  the  square  root  of 

the  denominator;  that  is,   A  7  = — -• 

:  b       V  6 

477.  'Rationalizing  the   denominator  simplifies  the  cal- 

rn  /~ 

culation ;   that  is,  -yj  —  is  more  difficult  than  -=-  • 

478.  When  the  same  number  occurs  as  the  radical  in 
a   series   of   terms,   the   terms  may  be   combined   by   the 
rule    for    adding    similar    terms.     This    usually    simplifies 
the  calculation. 

479.  The  theorem  of  Pythagoras  was  proved. 

480.  The  theorem  of  Pythagoras  furnishes  a  method  of 
constructing  the  square  root  of  a  number. 

481.  The     mean     proportional     construction     furnishes 
another  method  of  finding  the   square  root. 

482.  A  fractional  exponent  is  another  method  of  indicat- 
ing roots  and  powers;  thus,  x%  means  -$2?.   The  numerator 
indicates  the  power,  and  the  denominator  the  root. 

483.  a°  is  defined  as  1. 

484.  A  number  with  a  negative  exponent  is  defined  so 
as  to  be  equal  to  the  reciprocal  of  the  same  number  with 

a  positive  exponent;  that  is,  a~5  =  —  • 


INTERPRETATION  OF  ROOTS  AND  POWERS    423 

485.  The    cube   of    a   binomial  may   be   found   by   the 
following  formula: 

(a  +  6)3  =  a3  +  3  a26  +  3  a 


486.  Cube  roots  may  be  found  by  the  table,  graph,  or 
more  easily  by  logarithms  and  the  slide  rule  (the  last  two 
methods  will  be  shown  in  the  next  two  chapters). 


CHAPTER  XVII 

*  LOGARITHMS   APPLIED  TO   MULTIPLICATION,  DIVISION, 
ROOTS  AND  POWERS,  AND  VERBAL  PROBLEMS  INVOLV- 
ING EXPONENTIAL  EQUATIONS 

LOGARITHMS 

487.  Labor-saving  devices.  In  Chapter  IV  we  showed 
how  extensive  calculations  even  with  only  four  or  five 
place  numbers  are  apt  to  become  laborious  and,  in  some 
cases,  inaccurate  and  involving  unnecessary  steps.  We 
showed  how  to  minimize  the  inaccuracy  and  how  some  of 
the  unnecessary  steps  may  be  eliminated,  especially  with 
regard  to  the  processes  of  multiplication  and  division  by 
the  so-called  "  abbreviated  method."  But  in  many  cases 
the  work  remains  long  and  tedious,  even  with  the  use  of 
these  abbreviated  methods. 

In  Art.  449  will  be  found  a  table  of  powers  and  roots 
which  are  given  for  the  purpose  of  saving  time  and  labor. 
Scientific  books  include  similar  tables  which  help  to  save  time 
and  conserve  our  energy.  Other  labor-saving  devices  com- 
monly used  are  adding  machines,  cash  registers,  graphs,  etc. 

One  of  the  greatest  labor-saving  devices  by  which  diffi- 
cult problems  may  be  readily  solved  is  the  method  of 
calculation  by  logarithms.  This  chapter  will  be  devoted 
to  a  simple  explanation  of  this  method.  If  the  student 
will  study  the  chapter  carefully  and  solve  the  problems 
correctly,  he  will  get  a  foundation  in  logarithmic  work 
that  will  be  very  helpful  in  subsequent  work. 

424 


LOGARITHMS  425 

488.  Two    methods    of    multiplying.     We    are    already 
familiar  with  the  two  methods  of  multiplying  illustrated 
by  the  following  examples: 

100  x  1000  =  100,000. 

102  x  108  =  105 

=  100,000. 

The  student  will  observe  that  the  product  is  the  same 
in  both  examples,  but  that  the  method  used  in  the  second 
has  reduced  the  operation  of  actually  multiplying  the  two 
numbers  to  a  simple  problem  in  the  addition  of  exponents. 

Although  the  numbers  multiplied  here  are  in  each  case 
powers  of  10,  the  method  will  hold  for  other  bases  as  well. 
However,  we  shall  consider  only  the  base  10  in  our  sub- 
sequent discussion  since  it  is  the  one  commonly  used. 

Find  the  product  of  the  following  pairs  of  numbers  by 
the  two  methods  given  above : 

10  and  100.  10,000  and  100,000. 

1000  and  1000.  1000  and  1,000,000. 

489.  Powers  of  10.    From  the  preceding  exercises  it  is 
clear  that  we  can  multiply  together  numbers  which  are 
integral  powers  of  10  merely  by  adding  the  exponents  of 
these  powers.   Since  every  positive  number  may  be  expressed 
exactly  or  approximately  as  a  power  of  10,  we  may  obtain 
the  product  of  any  two  numbers  in  a  similar  manner  by 
adding  the  exponents  of  the  powers  of  10  which  equal  the 
respective  numbers.    For  example,  we  may  multiply  17.782 
and  3.162  by  adding  the  exponents  of  the  powers  of  10, 
which  equal  17.782  and  3.162  respectively. 

This  raises  two  important  questions:  (1)  What  powers 
of  10  equal  17.782  and  3.162?  (2)  What  is  the  value 
of  10  when  raised  to  the  sum  of  these  two  powers  ?  It  is 


426  GENERAL  MATHEMATICS 

possible  for  us  to  work  out  a  table  which  will  give  us  the 
powers  of  10  which  equal  17.782  and  3.162.  The  table 
below  shows  how  the  different  values  are  obtained.  The 
table  is  not  complete,  as  will  be  shown  later,  but  it  contains 
the  approximate  values  of  several  integral  and  fractional 
powers  of  10  which  we  need  at  this  point. 

We  know  that  10°  =  1,  101  =  10, 102  =  100, 103  =  1000,  and  so  on. 
\Vr  can  find  the  value  of  10°-5  as  follows  : 

10°-5  =  10*  =  VlO  =  3.162  (approx.). 

From  these  values  the  other  values  in  the  table  are  easily  found, 
as  the  student  can  verify. 

TABLE  OF  POWERS  OF  10 
10'  =    1.0000 


I  =  V8.162  =  1.7782 
10°-5  =  10z  =  VlO  =  3.1623 
100.75  _  (ioi.5)2  =  v'31.62  =  5.6234 
101  =  10.0000 

lQi-25  =  lOi  .  100.25  =    i7.78o 

lQi-5    _  i0i  .  100.5  _    31.623 

lOi-™  =10*  •  10<>-'5  =    56.234 

10*  =  100.000 

102.25  _  i0i .  iQi.25  _    177.82 

102-5   =  lOi  .  101.5  _    316.23 

jO-2.75  _  101  .  101-75  _      562.34 

10«  =1000.00 

We  may  now  resume  the  solution  of  the  problem  proposed 
above,  namely,  multiplying  17.782  by  3.162.  We  can  see 
by  referring  to  the  table  that  17.782  =  101-25  and  that 
3.162  =10°-5.  Hence,  17.782  x  3.162  =  101-25  xlO°-5  =  101-78, 
which,  by  referring  to  the  table,  we  see  is  equal  to  56.234 
(this  product  is  accurate  to  the  second  decimal  place). 


LOGARITHMS  427 

EXERCISE 

Check  the  preceding  result  by  actually  multiplying  17.782 
by  3.162,  and  account  for  the  difference  in  results.  Is  there  a 
significant  difference  ?  Is  the  result  obtained  by  actual  multi- 
plication accurate  to  more  than  two  decimal  places  ? 

490.  Logarithms ;  notation  for  logarithms.    In  the  equa- 
tion 17.782  =101-25   the    exponent   1.25  (which    indicates 
the  power  to  which  10  must  be  raised  to  give  17.782)  is 
called  the  logarithm  of  17.782  to  the  base  10. 

Thus  the  logarithm  of  a  number  to  the  base  10  is  tlie 
exponent  of  the  power  to  which  10  must  be  raised  to  equal 
that  number.  From  now  on  we  shall  assume  that  the  base 
is  10  when  we  speak  of  the  logarithm  of  a  number.  The 
symbol  for  logarithm  is  log.  Thus,  log  1000  =  3  is  read  "  the 
logarithm  of  1000  equals  3,"  the  base  10  being  understood. 

EXERCISE 

By  means  of  the  table  of  powers  of  10  in  Art.  489  find  log  1 ; 
log  10  ;•  log  100  ;  log  1.78;  log  316.23. 

491.  A  logarithm  is  an  exponent.    The  student  will  do 
well  to  remember  the  two  ways  of  thinking  of  an  expo- 
nent;   for  example,  in  the  equation  102  =  100  the  2  can  be 
thought  of  (a)  as  the  exponent  of  the  power  to  which  10 
must  be  raised  to  equal  100  ;  that  is,  102  =  100  ;  (b)  as  the 
logarithm  of  100  to  the  base  10;  that  is,  2  =  Iog10100. 

EXERCISES 

Read  the  following  in  two  ways : 

1.  101  =  10.     2.  102  =  100.     3.  108  =  1000.     4.  104  =  10,000. 


428  GENERAL  MATHEMATICS 

492.  Characteristic ;  mantissa.  A  glance  at  the  table  of 
Art.  489  will  show  that  each  exponent  of  10  (each  loga- 
rithm of  the  corresponding  numbers  to  the  right)  may  con- 
tain an  integral  part  and  a  fractional  part.  For  example,  in 
the  equation  101-25  =  17.782  the  1.25  (that  is,  log  17. 782) 
has  1  for  its  integral  part  and  0.25  for  its  decimal  (frac- 
tional) part.  In  102  =  100  the  entire  logarithm  is  integral. 
(Why?)  The  integral  part  of  a  logarithm  is  called  the 
characteristic  of  the  logarithm,  and  the  decimal  part  is 
called  the  mantissa  of  the  logarithm. 

The  characteristic  of  a  logarithm  of  any  number  can 
always  be  determined  at  sight.  For  example: 

log  10  =  1,  because  101  =  10 ; 

log  100  =  2,  because  102  =  100 ; 
log  1000  =  3,  because  103  =  1000 ; 

and  so  on.  But  these  numbers  are  all  integral  powers  of 
10.  However,  the  characteristic  of  the  logarithm  of  any 
other  number  may  be  obtained  as  well  by  observing  what 
powers  of  10  inclose  it.  For  example,  the  characteristic 
of  log  525  is  2  because  525  lies  between  the  second  and 
the  third  powers  of  10 ;  that  is,  between  102  =  100  and 
103  =  1000  (see  the  table,  Art.  489). 

It  is  not  so  easy  to  determine  the  mantissas  (the  deci- 
mal part)  of  the  logarithms  of  numbers.  We  have  worked 
out  a  few  of  these  in  the  table  of  Art.  489,  but  to  compute 
the  mantissas  for  all  other  numbers  in  this  way  would  be 
a  tedious'task.  Moreover,  the  methods  necessary  to  com- 
pute them  would  be  beyond  us  in  difficulty.  However, 
these  mantissas  have  been  computed  for  all  the  various 
powers  of  10  (by  more  advanced  methods),  and  they  appear 
in  the  table  of  mantissas  which  follows.  So  that  now  when 


LOGARITHMS  429 

we  want  to  know  what  the  logarithm  of  any  number  is, 
we  decide  (by  inspection)  what  the  characteristic  is  and 
then  look  in  the  table  for  the  mantissa. 

EXERCISES 

1.  Look  in  the  table  (pp.  430-431)  for  the  decimal  part  of 
the  logarithms  of  10  ;  15  ;  20  ;  38  ;  86  ;  "99. 

2.  What  is  the  decimal  part  of  the  logarithm  of  100  ? 

3.  What  is  the  power  to  which  10  must  be  raised  to  produce 
10,000  ? 

4.  What,  then,  is  the  logarithm  of  10,000  to  the  base  10  ? 

5.  Examine  the  table  carefully  and  tell  what  numbers  have 
integers  for  logarithms  ;  that  is,  those  that  have  zero  mantissas. 

6.  When  will  a  logarithm  have  a  decimal  mantissa  ? 

7.  Find  the  product  of  48  and  55. 
Solution.   By  means  of  the  table  we  see  that 

48  =  101-6812, 
and  that  55  =  101-7404. 

Therefore  48  x  55  =  101-6812  x  101-7404 


The  3  in  the  exponent  3.4216  tells  us  that  the  product  of  48  x  55 
is  a  number  between  the  3d  and  4th  powers  of  10  ;  that  is,  the  3 
tells  us  where  to  put  the  decimal  point.  We  must  find  the  mantissa 
0.4216  in  the  table  of  logarithms  and  see  what  number  corre- 
sponds to  it. 

If  we  look  in  the  table  of  mantissas  we  find  that  0.4216  is  the 
mantissa  of  the  logarithm  of  the  number  264.  Now  since  the  charac- 
teristic of  the  logarithm  is  3,  the  number  must  be  between  the  3d 
and  4th  powers  of  10  ;  that  is,  between  1000  and  10,000.  This  means 
that  the  decimal  point  comes  after  the  fourth  place,  so  that  we  must 
add  a  cipher  to  264.  Hence  the  number  is  2640. 


430 


GENERAL  MATHEMATICS 


TABLE  OF  MANTISSAS 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 
12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 

1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 

1614 

0253 
0645 
1004 
1335 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1072 
1399 
1703 

0374 
0755 
1106 
1430 
1732 

15 

16 

17 
18 
19 

1761 
2041 
2304 
2553 
2788 

1790 
2068 
2330 
2577 
2810 

1818 
2095' 
2355 
2601 
2833 

1847 
2122 
2380 
2625 
2856 

1875 
2148 
2405 
2648 

2878 

1903 
2175 
2430 
2672 
2900 

1931 
2201 
2455 
2695 
2923 

1959 
2227 
2480 
2718 
2945 

1987 
2253 
2504 
2742 
2967 

2014 
2279 
2529 
2765 
2989 

20 

21 
22 
23 
24 

3010 
3222 
,3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 
3838 

3075 
3284 
3483 
3674 
3856 

3(196 
3304 
3502 
3692 

3874 

3118 
3324 
3522 
3-711 
3892 

3139 
3345 
3541 
3729 
3909 

3160 
3365 
3560 

3747 
3927 

3181 
3385 
3579 
3766 
3945 

3201 
3404 
3598 
3784 

3962 

25 

26 
27 
28 
29 

3979 
4150 
4314 
4472 
4624 

3997 
4166 
4330 
4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216  • 
4378 
4533 
4683 

4065 
4232 
4393 
4548 
4698 

4082 
4249 
4409 
4564 
4713 

4099 
4265 
4425 
4579 
4728 

4116 
4281 
4440 
4594 
4742 

4133 
4298 
4456 
4609 
4757 

30 

31 

32 
33 
34 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
5366 

4843 
4983 
5119 
5250 
5378 

4857 
4997 
5132 
5263 
5391 

4871 
5011 
5145 
5276 
5403 

4886 
5024 
5159 
5289 
5416 

4900 
5038 
5172 
5302 

5428 

35 

5441 
5l6T 
5682 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 
5587 
5705 
5821 
5933 

5478 
5599 
5717 
5832 
5944 

5490 
5611 
5729 
5843 
5955 

5502 
5623 
5740 
5855 
5966 

5514 
5635 
5752 
5866 
5977 

5527 
5647 
5763 
5877 
5988 

5539 
5658 
5775 
5888 
5999 

5551 
5670 
5786 
5899 
6010 

"36 
37 
38 
39 

40 

41 

42 
43 
44 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 

6444 

6042 
6149 
6253 
6355 
6454 

6053 
6160 
6263 
6365 
6464 

6064 
6170 
6274 
6375 

6474 

6075 
6180 
6284 
6385 
6484 

6085 
6191 
6294 
6395 
6493 

6096 
6201 
6304 
6405 
6503 

6107 
6212 
6314 
6415 
6513 

6117 
6222 
6325 
6425 
6522 

45 

46 
47 
48 

49 

6532 
6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6561 
6656 

6749 
6839 

6928 

6571 
6665 
6758 
6848 
6937 

6580 
6675 
6767 
6857 
6946 

6590 
6684 
6776 
6866 
6955 

6599 
6693 
6785 
6875 
6964 

6609 
6702 
6794 
6884 
6972 

6618 
6712 
6803 
6893 
6981 

50 

51 
52 
53 
54 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 
7185 
7267 
7348 

7024 
7110 
7193 
7275 
7356 

7033 
7118 
7202 
7284 
7364 

7042 
7126 
7210 
7292 
7372 

7050 
7135 
7218 
7300 
7380 

7059 
7143 
7226 
7308 
7388 

7067 
7152 
7235 
7316 
73% 

No. 

0 

1 

2 

3 

4. 

5 

6 

7 

8 

9 

LOGARITHMS 


431 


TABLE  OF  MANTISSAS 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

56 
57 
58 
59 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 

7731 

7435 
7513 
7589 
7664 

7738 

7443 
7520 
7597 
7672 
7745 

7451 

7528 
7604 
7679 
7752 

7459 
7536 
7612 

7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 
7627 
7701 
7774 

60 

61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 
7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 

8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 

71 
72 
73 
74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8482_, 

'8488 
8549 
8609 
8669 
8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

854~3 
8603 
8663 

8722 

75 

76 

77 
78 
79 

8751 
8808- 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 
9025 

80 

81 
82 
83 

84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 

9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 
9284 

9079 
9133 
9186 
9238 
9289 

85 

86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9304 
9355 
9405 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 
9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

90 

91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 

9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 

9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

432  GENERAL  MATHEMATICS 

The  preceding  work  may  be  briefly  arranged  as  follows  : 
Let  N=  48x55 

log  48  =  1.6812 

log  55  =  1.7404 

Then  logA7  =  3.4216 

By  the  table,  N  =  2640. 

493.  Logarithm  of  a  product.  The  discussion  and  ex- 
amples in  Art.  492  have  shown  that  to  the  problem  of 
multiplying  two  powers  of  10  there  corresponds  the  problem 
of  adding  their  logarithms  (exponents).  This  may  be  stated 
briefly  as  the  first  law  thus:  The  logarithm  of  the  product 
of  two  numbers  is  the  sum  of  the  logarithms  of  the  factors; 
or,  by  formula,  log  (aby  =  log  a  -f-  log  b. 

It  is  easily  shown  that  the  law  also  holds  for  any  number 
of  factors  in  a  product  ;  that  is,  log  abc  =  log  a  +  log  b  +  log  c, 
and  so  on. 

EXERCISES 

1.  Check  by  actual  multiplication  the  logarithmic  method  of 
finding  the  product  of  48  and  55. 

2.  Find  by  means  of  logarithms  the  products  of  the  follow- 
ing numbers  : 

(a)  10  x  100  x  1000.          (b)  51  x  40.          (c)  83  x  6  x  2. 

3.  Find  by  using  logarithms  the  area  of  a  triangular  garden 
plot  whose  base  is  38  ft.  and  whose  altitude  is  17  ft. 

Solution.    The  formula  for  the  area  of  any  triangle  is 


Hence,  in  this  case,  A  =  19  •  17. 

log  19  =  1.2788 
log  17  =  1.2304 

Then  log  A  =  2.5092 

By  the  table,  A  =  323. 


LOGARITHMS 

4.   Solve  the  following  problems  by  means  of  logarithms  : 
4  (a)  What  is  the  area  of  a  rectangle   whose  base  is  70  and 
whose  altitude  is  32  ? 

(b)  What  is  the  area  of  a  parallelogram  whose  base  is  64  and 
whose  altitude  is  85  ? 

(c)  What  is  the  area  of  a  square,  a  side  of  which  is  29  ? 

Solution.   The  formula  for  the  area  of  a  square  is 

A  =  ,s  •  *•  =  *•-. 

In  this  case  .1  =  29  x  29  =  292. 

log  29  =  1.4624 
log  29  =  1.4624 

Therefore  log  A  =  2.9248 

and,  by  the  table,  A  =841. 

494.  Logarithm  of  a  power.  Obviously  the  result  in 
Ex.  4,  (c),  would  have  been  the  same  if  we  had  multiplied 
the  logarithm  of  29  by  2  instead  of  adding  it  to  itself. 
In  like  manner,  if  we  want  the  logarithm  of  503,  we  can 
obtain  it  either  by  using  the  logarithm  of  50  three  times 
as  an  addend  or  by  taking  three  times  the  logarithm  of  50. 
Thus,  log  503  =  3  log  50  =  3  x  1.6990  =  5.0970. 

This  discussion  illustrates  a  very  important  law  ;  namely, 
that  the  logarithm  of  a  number  roiled  to  a  power  in  the  exponent 
of  the  power  times  the  logarithm  of  the  number ;  or,  by  for- 
mula, log  a"  =  n  log  a. 

EXERCISES 

1.  Find  by  the  preceding  method  the  logarithms  of  the 
following  numbers:  222;  II8;  154. 

2.  Find  by  logarithms  the  volume  of  a  cubical  cake  of  ice 
one  edge  of  which  is  10  in. ;  11  in. 

3.  The  area  .1  of  a  circle  is  given  by  the  formula  A  =  irr, 
where  /•  is  the  radius  of  the  circle.    What  is  the  area  of  a  circle 
whose  radius  is  6  in.  ?  (Use  logTr  =  0.4971.) 


434  GENERAL   MATHEMATICS 

495.  Logarithm  of  a  quotient.  The  method  of  computing 
by  logarithms  is  as  useful  in  division  as  in  multiplication. 
In  order  to  make  the  method  clear,  let  us  review  our  two 
methods  of  dividing  one  number  by  another. 

(a)  -          -  =  1000,  by  actual  division. 

(b)  -          —  =  —  -2=  103=  1000,  by  subtracting  the  exponents. 

1UU  1U 

Here,  as  in  multiplication,  we  obtain  the  same  result 
by  either  method,  but  the  second  method  reduces  the 
operation  of  actual  -division  to  a  simple  problem  of  sub- 
tracting exponents. 

EXERCISES 

1.  Find  the  following  quotients  by  the  two  methods  just 
discussed  : 

100,000  1,000,000 

1000    '  10,000 

2.  Divide  100  by  31.623  by  using  the  table  of  Art.  489. 

So"*ion-        = 


XOTE.    The  student  should  check  this  result  by  actual  division. 
3.  Divide  562.34  by  31.62  by  using  the  table  of  Art.  489. 

We  may  obtain  the  quotient  of  any  two  numbers  in 
like  manner  by  subtracting  the  exponent  of  the  power  of 
10  equal  to  the  divisor  from  the  exponent  of  the  power  of 
10  equal  to  the  dividend.  Keeping  in  mind  the  definition 
of  a  logarithm  it  is  clear  that  this  fact  may  be  expressed  as 
a  law  ;  thus,  the  logarithm  of  the  quotient  of  two  numbers  is 
the  logarithm  of  the  dividend  minus  the  logarithm  of  the 

divisor  ;  or,  as  a  formula,  log  IT)—  l°9  a  ~  1°9  ^ 


LOGAEITHMS  435 

EXERCISES 

1.  Given  log  2  =  0.3010,  log  3  =  0.4771,  find  log  f  ;  log  J 

2.  Find    the    value    of    the    following    fractions    to    three 
significant  figures  by  using  logarithms  : 

x  59  x  85  .  381  x  II3 

<a>  —43"  ~^~ 

.   752  x  350  71  x  48  x  253 

-~  ~" 


HINT.  To  find  the  logarfthm  of  each  fraction,  arid  the  logarithms 
of  the  factors  of  the  numerator  and  from  this  sum  subtract  the  sum 
of  the  logarithms  of  the  factors  of  the  denominator. 

496.  Position  of  the  decimal  point.  Since  the  multipli- 
cation or  division  of  a  number  by  10  amounts  to  a  moving 
of  the  decimal  point  one  place  to  the  right  in  multiplication 
and  one  place  to  the  left  in  division,  and  since  the  multi- 
plication or  division  of  a  number  by  100  amounts  to 
moving  the  decimal  point  two  places  to  the  right  or  left, 
and  so  on,  the  position  of  the  decimal  point  in  a  number 
whose  logarithm  we  are  seeking  affects  the  characteristic  onli/. 

The  truth  of  the  foregoing  statement  can  be  seen  best  by 
means  of  the  table  in  Art.  489.  In  this  table,  for  example, 

!  Oo.25  =1.7782,  or  log  1.7782  =  0.25. 
If  we  multiply  both  sides  of  this  equation  by  10,  we  get 

101-2B=  17.782,  or  log  17.782  =  1.25. 

Again,  multiplying  both  sides  of  this  last  equation  by  10, 
1  02.25  =  177.82,  or  log  177.82  =  2.25, 

and  so  on.  The  student  will  observe  that  only  the  integral 
part  of  the  exponent  of  10  (the  logarithm  of  the  number) 
is  changing,  and  that  the  array  of  figures  remains  the 


436  GENERAL  MATHEMATICS 

same  even  though  the  decimal  point  moves  one  place  to 
the  right  after  each  multiplication.  In  like  manner,  if  we 
divide  both  sides  of  the  equation  100-25  =  1.778  by  10,  and 
continue  the  division,  we  obtain 

loo.*- 1  =  0.1778,  or  log  0.1778  =  0.25  - 1; 
100.25-2  =  0.01778,  or  log  0.01778  =  0.25  -  2, 
lOo.ffi-3  =  0.001778,  or  log  0.001778  =  0.25  -  3, 

and  so  on. 

The  logarithms  0.25-1,  0.25-2,  0.25-3,  etc.  are 
negative  quantities,  but  they  are  not  in  the  form  in  which 
.we  usually  write  negative  numbers.  However,  if  we  adopt 
these  forms,  the  mantissas  of  all  our  logarithms  will  not 
only  be  positive  but  they  will  be  the  same  for  the  same 
array  of  figures  no  matter  where  the  decimal  point  is  found. 

Thus  the  mantissa  of  log  1.778  is  the  same  as  the 
mantissa  of  log  0.001778,  as  was  shown  above.  These  two 
logarithms  differ,  therefore,  only  in  their  characteristics. 
In  some  texts  log  0.1778  is  written  1.25  instead  of  0.25  —  1. 
To  agree  with  this  statement  the  dash  above  the  1  means 
that  only  the  1  is  negative.  Some  books  prefer  the  form 
9.25-10  instead  of  0.25-1  or  1.25.  The  student  can 
easily  see  that  9.25—10  has  the  same  value  as  0.25—1. 
\Ve  shall  later  see  another  advantage  of  the  form  9.25  —  1 0. 

The  preceding  discussion  may  be  summarized  in  two 
statements  : 

1.  We  agree  to  express  the  logarithm  of  any  number  in  a 
for tn    such    that   its   mantissa    shall   be  positive.     This   can 
always  be  done,  whether  the  number  is  greater  or  less  than 
unity.    In  either  case  the  positiveness  or  negativeness  of 
the  number  is  shown  entirely  by  means  of  the  characteristic. 

2.  Two  numbers  containing  the  same  succession  of  digits, 
that   ?X   differing  only  in  the  position  of  the  decimal 


LOGARITHMS  437 

will  have  logarithms  that  differ  only  in  the  value  of  the 
characteristic.  This  explains  why  we  called  the  table  a 
table  of  mantissas  and  why,  in  looking  up  the  logarithm 
of  a  number,  we  need  pay  no  attention  to  the  decimal 
point  in  the  number.  The  same  table  of  mantissas  serves 
both  for  numbers  greater  and  less  than  unity. 

497.  Table   of   characteristics.    The    following  'table   is 
given   in    order   that   the    student    may   determine    more 
quickly  the  characteristic  of  the  logarithm  of  any  number 
between  10  to  the  minus  6th  power  and  10  to  the  plus 
7th   power.     This   is    about   as   much   of   a   range    as  we 
shall  ever  need.    The  table  can  be  extended  upward  or 
downward  at  will. 

TABLE  OF  CHARACTERISTICS 

10~6  =  0.000001         10J=10 

10-5-0.00001        io2=ioo 

10-4-0.0001  103  =  1,000 

io-*  =  o.ooi  io4= 10,000 

10-2-0.01.'  105  =  100,000 

10- !  =  0.1  io6 =1,000,000 

10°  =  1  IO7- 10,000,000 

For  example,  if  we  want  the  logarithm  of  2142,  we 
know  that  its  characteristic  is  3,  because  2142  lies  between 
the  3d  and  4th  powers  of  10.  Again,  if  we  want  the 
logarithm  of  0.0142,  we  know  that  the  characteristic  of 
the  logarithm  •  is  —2,  because  0.0142  lies  between  the 
minus  2d  and  the  minus  1st  power  of  10. 

498.  Interpolation.    So  far  we  have  shown  the  student 
how  to  find  the  logarithms  of  three-digit  numbers  only.    In 
order  to  be  able  to'  find  the  logarithms  of  numbers  of  more 
than  three  digits,  and  to  find  the  numbers  corresponding 


438  GENERAL  MATHEMATICS 

to  various  logarithms  which  we  may  obtain  in  calculation, 
it  is  necessary  for  us  to  learn  how  to  make  further  use  of 
the  table  in  Art.  492.  We  shall  proceed  to  consider  two 
typical  examples. 

1.  Find  the  logarithm  of  231.6. 

Solution.  From  the  table  of  Art.  497  it  is  clear  that  the  char- 
acteristic of  the  logarithm  is  2.  To  find  the  mantissa,  we  ignore  the 
decimal  point  and  look  in  the  table  of  Art.  492  for  the  mantissa 
of  2316.  Reading  down  the  left-hand  column  of  the  table,  headed 
"  No.,"  we  find  23.  The  numbers  in  the  same  horizontal  row  with  23 
are  the  mantissas  of  the  logarithms  230,  231,  232,  and  so  on. 

We  want  to  find  the  logarithm  of  2316.    We  can  now  write 

log  232  =  2.3655 

log  231  =  2.3636 

Tabular  difference  =  0.0019 

Now  since  231.6  is  f'6  of  the  way  from  231  to  232,  we  add  {'^ 
of  the  tabular  difference  0.0010  to  the  logarithm  of  231.  Thus, 

log  231.6  =  2.3636  +  ^  x  0.0019. 
Therefore  log  231.6  =  2.3647. 

The  process  of  obtaining  the  logarithm  of  a  number 
in  this  way  is  called  interpolation.  The  student  should 
practice  this  method  by  finding  the  logarithms  of  several 
four-digit  numbers. 

2.  Find  the  number  whose  logarithm  is  0.3883  —  1. 

Solution.  We  know  at  once  that  the  number  is  a  decimal  fraction 
lying  between  the  minus  1st  and  the  0  power  of  10.  This  tells  us 
that  the  decimal  point  comes  just  before  the  first  significant  figure 
in  the  number.  If  we  look  in  the  table  of  Art.  492  we  do  not  find 
the  mantissa  3883,  but  we  find  3892,  which  is  a  little  greater,  and 
3874,  which  is  a  little  less ;  that  is, 

0.3892  -  1  =  log  0.245. 
0.3874  -  1  =  log  0.244. 

Since  0.3883  —  1  is  the  logarithm  of  the  number  we  want,  the 
number  lies  between  0.244  and  0.245.  Now  0.3883  - 1  is  T98  of  the 


LOGARITHMS       .  439 

way  from  log  0.244  to  log  0.245  ;  hence  the  number  corresponding 
to  the  logarithm  0.3883  -  1  lies  ^,  or  i,  of  the  way  from  0.244  to 
0.245.  Therefore  the  number  desired  is  0.2445. 

Here  the  process  of  interpolation  is  used  on  the  inverse 
problem,  that  of  finding  a  number  when  its  logarithm  is 
given. 

EXERCISES 

1.  Find  the  logarithms  of  the  following  numbers:  745;  83.2; 
91200;  0.567;  0.00741.    (No  interpolation.) 

2.  Find  the  logarithms  of  the  following  numbers  :  6542  ; 
783.4;  91243;  0.4826;  0.002143.    (Interpolation.) 

3.  Find  the  numbers  whose  logarithms  are  2.6075  :  1.4249; 
0.3054;  0.0212-2;  0.8457-1.    (No  interpolation.) 

4.  Find  the  numbers  whose  logarithms  are  2.3080;  1.936; 
0.8770  ;  0.0878  -  2.    (Interpolation.) 

499.  Extraction  of  roots  by  means  of  logarithms.  In 
Art.  460  we  discussed  the  meaning  of  fractional  exponents 
and  showed  that 

a*  =  Va  ;  «*  =  Va  ;  a*  =  V«  ;  etc. 

If  we  assume  that  the  theorem  of  Art.  494,  regarding 
the  logarithm  of  a  number  raised  to  a  power,  holds  for 
fractional  exponents,  then 

log  Va  =  log  a*  =  -|  log  a, 
log  v  a  =  log  a  *  =  1  log  a, 

and  so  on.  This  illustrates  the  truth  of  another  theorem, 
namely,  that  the  logarithm  of  any  root  of  a  number  is  equal 
to  the  logarithm  of  the  number  divided  by  the  index  of  the  root. 


9 

Thus,      log  V542  =  J  log  542  =  ^~-  =  1.3670. 

Now  1.3670  is  the  logarithm  of  23.28  -.    Therefore  the 
square  root  of  542  is  approximately  23.28  —  . 


440  GENERAL  MATHEMATICS 

If   the  logarithm  of   the  number  is  negative,  the  root 
may  be  found  as  in  the  following  examples. 
Find  by  logarithms : 

(a)   V0.472.  (b)   ^0.472.  (c)  -V/O472. 

Solution.  Log  0.472  =  0.6739  —  1.  Now  if  we  attempt  to  take  ^ 
of  this  negative  logarithm,  we  shall  obtain  a  fractional  characteristic 
that  would  be  confusing.  Therefore,  in  order  to  make  it  possible 
to  keep  the  mantissa  positive  and  the  characteristic  (after  the 
division)  an  integer,  we  write 

log  0,172  =  1!).<>73!>  -20, 

a  number  which  the  student  can  readily  see  is  equal  to  0.6739  —  1, 
and  which  has  the  added  advantage  referred  to  above.    We  now  get 

(a)  log  Vo.472  .=  *  (19.6739  -  20)  =  9.8369  -  10. 
In  like  manner, 


(b)  log  V 0.472  =  i  (29.6739  -  30)  =  9.8913  -  10, 
and  («)  log  v  0.472  =  J  (39.6739  -  40)  =  9.9185  -  10. 


In  (a),  above,  log  Vo.472  =  9.8369  -10.  This  means  that 
the  characteristic  is  —1  and  that  the  mantissa  is  0.8369. 
By  reference  to  the  table,  we  find  that  0.8369  is  the  man- 
tissa of  the  logarithm  of  687.  Since  the  characteristic  is 
- 1,  the  number  lies  between  the  minus  1st  and  the  0  power 
of  10  ;  hence  the  decimal  point  comes  just  before  the  6,  and 
V0.472  =  0.687.  The  student  should  check  this  result  by 
actually  extracting  the  square  root  of  0.472  by  the  method 
given  in  Art.  446. 

EXERCISES 

1.  Find  by  logarithms  : 

V9604 ;   V153.76 ;   V0.000529 ;   A/10648 ; 
^42§75 ;  -v/3.375 ;  ^0.001728. 

2.  Given  a  =  4.25,  ft  =  22.1,  and  c  =  0.05,  find  by  logarithms 

•  I      72 

the  value  of  \|—  to  three  significant  figures. 


LOGARITHMS  441 


.     182.41  '    ,    3)2.42x35.1 

3.  Find  by  logarithms  the  value  of  V  and  -\\ ~^r^ — T^ 

i  4.oo^  i   i£>.£  X  J." 

to  three  significant  hgures. 

4.  The  velocity  v  of  a  body  that  has  fallen  a  distance  of 
,s-  feet  is  given  by  the  formula  v  =  V2  </.->•.    If  g  =  32.16,  what 
is  the  velocity  acquired  by  a  body  in  falling  30  ft.? 

5.  If  a  bullet  is  shot  upwards  with  an  initial  velocity  of  v, 
the  height  .s  to  which  it  will  rise  is  given   by  the  equation 
r  =  V2  gs.    What  must  be  the  minimum  velocity  at  the  start 
of  a  bullet  fired  upwards  if  it  is  to  strike  a  Zeppelin  1500  ft. 
high?    (Assume  y  =  32.16.) 

6.  The  time  t  of  oscillation  of  a  pendulum  /  centimeters  long 

is  given  by  the  formula  t  =  7r\|  — j-  •   Kind  the  time  of  oscilla- 

i  j)oO 

tion  of  a  pendulum  78.22  centimeters  long. 

*  7.  If  the  time  of  oscillation  of  a  pendulum  is  1  sec.,  what 
is  its  length  ? 

*8.  The  area  of  any  triangle  is  given  by  the  formula 

A  =  Vs  (s  —  a)  (s  —  b}  (s  —  c), 

where  a,  b,  and  c  are  the  sides  of  the  triangle,  and  s  equals 

a  +  b  4-  c 

one  halt  the  perimeter,  or 

& 

Using  the  preceding  formula,  find  by  means  of  logarithms  the 
number  of  acres  in  a  triangular  field  whose  sides  are  15,  38.2, 
and  45.3  rods  respectively.  (1  A.  =  160  sq.  rd.) 

*9.  The  area  A  of  the  cross  section  of  a  chimney  in  square 
feet  required  to  carry  off  the  smoke  is  given  by  the  formula 

0.6  P 

—      /~  ' 

V// 

where  P  is  the  number  of  pounds  of  coal  burned  per  hour,  and 
h  is  the  height  of  the  chimney  in  feet.  Find  out  what  the  area 
of  a  cross  section  of  a  chimney  80  ft.  high  should  be  to  carry 
off  the  smoke  if  560  Ib.  of  coal  are  burned  per  hour. 


442 


GENERAL  MATHEMATICS 


*  10.  The  average  velocity  v  of  the  piston  head  in  inches  per 
second  for  a  steam  engine  is  given  by  the  formula 


where  s  denotes  the  distance  (in  inches)  over  which  the  piston 
head  moves,  and  p  is  the  number  of  pounds  of  pressure  of 
steam  in  the  cylinder.  Find  v  if  s  =  30.24  in.  and  p  =  115  Ib. 

*11.  In  the  equation  x  —W1J  what  is  the  value  of  x  when 
y  =  0  ?  when  y  —  1  ?  when  y  =  2  ?  when  ?/=  —  !? 

*12.  Fill  in  the  following  table  for  the  equation  x  =10?/. 


y 

0 

i 

2 

-1 

-  2 

-3 

i 

I 

i 

4 

5 

X 

i 

10 

100 

NOTE.  For  y  equaling  ^,  ^,  2^,  etc.,  use  the  table  of  mantissas, 
Art.  492. 

*  13.  Plot  the  results  in  the  table  of  Ex.  12  and  draw  the  curve, 
thus  obtaining  the'graph  for  x  =  lO2'  (or  y  —  log  x)  (Fig.  300). 


FIG.  300.    GRAPH  OF  x  = 


*14.  Show  that  the  graph  for  x  =10^  (Fig.  300)  makes  clear 
the  following  principles  : 

(a)  A  negative  number  does  not  have  a  real  number  for  its 
logarithm. 


LOGARITHMS  443 

(b)  The  logarithm  of  a  positive  number  is  positive  or  nega- 
tive according  as  the  number  is  greater  or  less  than  1. 

(c)  The  greater  the  value  of  x,  the  greater  its  logarithm. 

(d)  As  x  gets  smaller  and  smaller,  its  logarithm  decreases 
and  becomes  smaller  and  smaller. 

*15.  Find  by  the  graph  of  Ex.  13  the  logarithm  of  2.25; 
of  4.5 ;  of  1.1 ;  of  2.8. 

*J6.  Of  what  number  is  0.35  the  logarithm?  0.5?  0.42? 
*17.  Check  your  results  for  Exs.  15  and  16  by  the  results 
given  in  the  table  of  Art.  492. 

*  500.  Exponential  equations.  Instead  of  finding  the  log- 
arithm of  1000  to  the  base  10,  we  could  arrive  at  the  same 
result  by  solving  the  equation  10r  =  1000,  for  this  equa- 
tion asks  the  question,  What  power  of  10  equals  1000? 
In  other  words,  What  is  the  logarithm  of  1000  to  the 
base  10  ?  An  equation  like  this,  in  which  an  unknown  is 
involved  in  the  exponent,  is  called  an  exponential  equation. 

EXERCISE 

Give  five  examples  of  exponential  equations  where  10  is 
the  base. 

*501.  Method  of  solving  exponential  equations.  The  sim- 
plest exponential  equations  may  be  solved  by  inspection 
just  as  the  logarithms  of  many  numbers  can  be  given  by 
inspection.  Where  an  exponential  equation  cannot  be 
solved  readily  by  inspection,  logarithms  may  be  employed 
to  simplify  the  process.  We  will  illustrate  each  case. 

Solution  I  (by  inspection). 

(a)  If  2X  =    4,  then  x  =  2.  (e)  If  10?'  =       100,  then  y  -  2. 

(b)If3a;=    9,  then*  =  2.  (f)IflOj;=     1000,  then  x  -  3. 

(c)  If  2V  =    8,  then  »/  =  3.  (g)  If  10»  =  10,000,  then  y  =  4. 

(,1)  Tf  3*  =  81,  then  x  =  4. 


444  GENERAL  MATHEMATK  \S 

Solution  II  (by  using  logarithms). 

Solve  the  equation  2X  =  6  for  x. 
Taking  the  logarithms  of  both  sides : 

log  2X  =  log  6, 
or  x  log  2  =  log  6. 

.    z  =  !2£e  =  0™2  =  2  58 
log  2      0.3010 

1  G. 

The  student  must  remember  that  — ^— -  is  not  equal  to 

6  log2 

log  -•     The  first  is  a  fraction   obtained  by  dividing  one 

A 

logarithm  by  another,  and  involves  division ;  the  second 
indicates  that  the  logarithm  of  a  fraction  is  to  be  found, 
and  involves  subtraction. 

EXERCISE 

Solve  the  following  equations  for  x : 

(a)  2X  =  7.       (b)  3X  =  5.       (c)  4*  =  10.       (d)  (1.12)*  =  3. 

502.  Interest  problems  solved  by  logarithms.  Some  im- 
portant problems  in  interest  may  be  solved  by  means  of 
exponential  equations  and  logarithms.  The  following 
simple  example  will  illustrate  the  principle: 

In  how  many  years  will  a  sum  of  money  double  itself  at  6% 
if  the  interest  is  compounded  annually  ? 

Solution.  In  one  year  $1  will  amount  to  $1.06  ;  in  two  years  the 
amount  will  be  1.06  x  1.06,  or  (1.06)2 ;  in  three  years  the  amount 
will  be  (1.06)3;  and  so  on.  Therefore  in  x  years  the  amount  of  $1 
will  be  (1.06)3'.  Then,  if  the  money  is  to  double  itself  in  x  years, 
the  conditions  of  the  problem  will  be  represented  by  the  equation 
(1.06)'*  =  2.  Solving  this  equation  for  x,  we  get  x  =  12.3  (approx.). 
Therefore  a  sum  of  money  will  double  itself  at  6%  (compounded 
annually)  in  about  12.3  yr. 


LOGARITHMS  445 

EXERCISES 

1.  Explain  the  solution  of  the  problem  given  in  Art.  502. 

2.  If  the  interest  is  compounded  annually,  in  how  many  years 
will  a  sum  of  money  double  itself  at  3%  ?  3|-%  ?  4%  ?  5%  ? 

3.  In  how  many  years  will  a  sum  of  money  treble  itself  at 
4^  interest  compounded  annually  ?  semiannually  ? 

4.  The  amount  of  P  dollars  for  n  years  at  •/•%,  compounded 
annually,  is  given  by  the  formula  .1  =P(1  +/•)".     Find  the 
amount  of  $1200  for  10  yr.  at  4%. 

Solution.    In  this  problem  P  =  1200,  r  =  0.04,  n  =  10. 

^  =  1200(1  +  0.04)™ 
The  computation  may  be  arranged  as  follows : 

log  1200  =  3.0792 

10  log  1.04  =  0.1700 

log  A  =  3.2492 

Therefore  A  =  1775,  number  of  dollars  in  the 

amount. 

NOTE.  As  a  matter  of  fact,  this  value  of  A  is  not  exact,  because 
we  are  using  only  four-place  tables.  In  practice  the  value  of  the 
problem  should  determine  the  kind  of  tables  used.  The  greater  the 
number  of  places  given  in  tin'  tables  used,  the  greater  the  accuracy 
of  the  result. 

5.  What  will   $5000  amount   to  in   5  yr.  at  3%,   interest 
compounded  annually  ?  semiannually?  quarterly? 

6.  Approximately    three    hundred    years    ago    the    Dutch 
purchased  Manhattan  Island  from  the  Indians  for  $24.    What 
would  this  $24  amount  to  at  the  present  time  if  it  had  been 
placed  on  interest  at  6%' and  compounded  annually? 

7.  What  would   he  the  amount  of  10  at  the  present  time 
if  it  had  been  placed  on  3%  annual  compound  interest  tifty 
years  ago  ? 


446  GENERAL  MATHEMATICS 

8.  A  boy  deposited  300  in  a  savings  bank  on  3%  interest, 
the  interest  to  be  compounded  annually.    He  forgot  about  his 
deposit  until  fifteen  years  later,  when  he  found  the  receipt 
covering  the  original  deposit.   What  did  the  300  amount  to  in 
the  fifteen  years  ? 

9.  What  sum  will  amount  to  $1600  in  10  yr.  at  6%,  interest 
being  compounded  annually  ? 

10.  What  sum  will  amount  to  $  2500  in  5  yr.  at  3%,  interest 
being  compounded  annually  ? 

11.  In  how  many  years  will  $4000  amount  to  $8500  at  6%, 
interest  being  compounded  annually  ? 

12.  What  would  be  the   amount  to-day  of  1   cent  which 
nineteen  hundred  and  twenty  years  ago  was  placed  on  interest 
at  6%,  compounded  annually  ?    Find  the  radius  in  miles  of  a 
sphere  of  gold  which  has  this  value. 

NOTE.  A  cubic  foot  of  gold  weighs  1206  pounds  avoirdupois, 
one  pound  being  worth  approximately  $290. 

The  volume  of  a  sphere  is  given  by  the  formula  V  =  3  irr3,  where 
V  is  the  volume  and  r  the  radius  of  the  sphere. 

No  doubt  the  pupil  is  convinced  of  the  value  of  log- 
arithms as  a  labor-saving  device  in  complicated  arithmetic 
computations.  Since  he  will  meet  numerous  opportunities 
for  applications,  the  lists  of  problems  in  the  chapter  are 
brief,  the  aim  being  to  give  just  enough  illustrations  to 
make  clear  the  meaning  of  the  principles  involved. 

HISTORICAL  NOTE.  Logarithms  were  invented  by  John  Napier 
(1550-1617),  baron  of  Merchiston  in  Scotland.  His  greatest  purpose 
in  studying  mathematics  was  to  simplify  and  systematize  arithmetic, 
algebra,  and  trigonometry.  The  student  should  read  about  Xapier's 
"rods,"  or  "bones,"  which  he  designed  to  simplify  multiplication  and 
division  (Encyclopaedia  Britannica,  llth  ed.). 

It  was  his  earnest  desire  to  simplify  the  processes  that  led  him 
to  invent  logarithms ;  and,  strange  as  it  may  seem,  he  did  not  consider 
a  logarithm  as  an  exponent.  In  his  time  the  theory  of  exponents  was 


LOGARITHMS  447 

not  known.  A  Swiss  by  the  name  of  Jobst  Biirgi  (1552-1632)  may 
have  conceived  the  idea  of  logarithms  as  early  or  earlier  than 
Napier  and  quite  independently  of  him,  but  he  neglected  to  pub- 
lish his  results  until  after  Napier's  logarithms  were  known  all  over 
Europe. 

Henry  Briggs  (1561-1630),  who,  in  Napier's  time,  was  professor  of 
geometry  in  Gresham  College,  London,  became  very  much  interested 
in  Napier's  work  and  paid  him  a  visit.  It  is  related  that  upon 
Briggs's  arrival  he  and  Napier  stood  speechless,  observing  each  other 
for  almost  a  quarter  of  an  hour.  At  last  Briggs  spoke  as  follows : 
"  My  lord,  I  have  undertaken  this  long  journey  purposely  to  see  your 
person,  and  to  know  by  what  engine  of  wit  or  ingenuity  you  came 
first  to  think  of  this  most  excellent  help  in  astronomy,  namely,  the 
logarithms,  but,  my  lord,  being  by  you  found  out,  I  wyonder  nobody 
found  it  out  before,  when  now  known  it  is  so  easy." 

After  this  visit  Briggs  and  Napier  both  seem  to  have  seen  the 
usefulness  of  a  table  of  logarithms  to  the  base  10,  and  Briggs  devoted 
himself  to  the  construction  of  such  tables.  For  this  reason  logarithms 
to  the  base  10  are  often  called  Briggsian  logarithms. 

Abbott  says  that  when  Napoleon  had  a  few  moments  for  diversion, 
he  often  spent  them  over  a  book  of  logarithms,  which  he  always 
found  recreational. 

Miller  in*  his  "  Historical  Introduction  to  Mathematical  Litera- 
ture (p. 70)  says  :  "The  fact  that  these  logarithms  had  to  be  computed 
only  once  for  all  time  explains  their  great  value  to  the  intellectual 
world.  It  would  be  difficult  to  estimate  the  enormous  amount  of  time 
saved  by  astronomers  and  others  through  the  use  of  logarithm  tables 
alone."  (For  further  reading  see  Cajori's  "History  of  Elementary 
Mathematics,"  pp.  155-167.  Consult  also  the  New  International  Cy- 
clopedia, which  contains  a  great  deal  of  excellent  historical  material.) 

SUMMARY 

503.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases :  logarithm,  characteristic,  mantissa, 
interpolation,  and  exponential  equation. 

504.  The  theory  and  practical  value  of  logarithms  has 
been  discussed  in  as  elementary  a  way  as  possible  so  that 


448  GEN  KKA  L  MATHEMATK 'S 

the  student  may  be  able  to  appreciate  the  value  of  this 
powerful  labor-saving  device. 

505.  This  chapter  has  taught  the  student  four  impor- 
tant logarithmic  formulas  : 

(a)  log  ab  —  log  a  +  log  b*  (c)  log  au  =  n  log  a. 

(b)  log  -  =  log  a  —  log  b.  (d)  log  Va  =     °     • 

o  n 

506.  The  position   of  the  decimal  point  in  any  result 
depends  entirely  upon  the  characteristic  of  the  logarithm 
of  the  number  sought.     Thus,  two  numbers  having  the 
same   succession   of  digits  will  have  the  same  mantissa. 
The   mantissa  of   the  logarithm   of  a  number   is    always 
positive;  the  characteristic  may  be  either  +  or  — . 

507.  This  chapter  has  taught  methods  of  solving  loga- 
rithmic and  exponential  equations. 

508.  The  student  has  been  taught  how  to  solve  verbal 
problems  by  means  of  logarithms,  for  example,  the  interest 
problem. 


CHAPTER  XVIII 

THE  SLIDE  RULE  AND  GEOMETRIC  REPRESENTATION  OF 
LOGARITHMS  PRACTICALLY  APPLIED  TO  THE  SLIDE  RULE 

509.  General  description  of  the  slide  rule.  The  theory  of 
the  slide  rule  is  based  on  the  elementary  ideas  and  prin- 
ciples of  logarithms,  and  anyone  can  learn  to  use  the  slide 
rule  who  has  the  ability  to  read  a  graduated  scale. 

The  slide  rule  may  be  used  in  nearly  all  forms  of  cal- 
culation and  is  gradually  coming  into  general  use  in 
many  different  fields:  to  the  practicing  engineer  and  to 
the  student  of  the  mathematical  sciences  it  is  invaluable ; 
to  the  accountant  and  the  statistician  it  is  an  instru- 
ment of  great  service.  The  student  will  therefore  find  it 
advisable  to  learn  all  he  can  about  the  actual  use  of  this 
important  device. 

In  Chapter  II  we  discussed  the  graphical  method  of 
adding  and  subtracting  line  segments  as  an  interpretation 
of  the  addition  and  subtraction  of  numbers.  Since  to  a 
multiplication  of  two  numbers  there  corresponds  an  addi- 
tion of  their  logarithms,  it  is  possible  to  obtain  the  loga- 
rithm of  a  product  graphically  by  adding  line  segments 
whose  lengths  represent  the  logarithms  of  the  factors.  In 
order  to  carry  out  this  plan,  a  method  of  actually  finding 
a  line  segment  whose  length  shall  represent  the  logarithm 
of  a  given  number  has  been  developed. 

The  slide  rule  is  a  mechanical  device  for  determining 
products,  quotients,  powers,  and  roots  of  numbers  by 
graphical  addition  and  subtraction. 

449 


450 


GENERAL  MATHEMATICS 


Mantissas  of  logarithms  of  numbers  from  1  to  10  (which, 
as  we  have  seen,  are  the  same  for  numbers  from  100  to  1000 
or  from  1000  to  10,000)  are  laid  off  to  a  certain  scale  on 
two  pieces  of  rule  (see  Fig.  301)  which  are  made  to  slide  by 


nr 


FIG.  301.    SLIDE  KUL.K 

each  other  so  tliat  the  sums  or  differences  of  the  logarithms 
can  be  obtained  mechanically. 

The  scale  is  numbered  from  1  to  10  at  the  points  which 
mark  the  logarithms  of  the  several  numbers  used.  The  com- 
mon scale  is  5  in.  long  and  the  common  rule  10  in.  long,  so 
that  the  series  of  logarithms  is  put  on  twice,  and  the  number- 
ing either  repeated  for  the  second  set  or  continued  from  10 


A  1 

2             34              68 

B     1                      2              34 
C 

D 

FIG.  302 

to  100.  The  most  common  form  of  the  rule  is  the  Mannheim 
Slide  Rule.  On  this  rule,  which  is  made  essentially  as  shown 
in  Figs.  302  and  303,  there  are  two  scales  A  and  B  just  alike 
(A  on  the  rule  and  B  on  the  slide),  and  two  other  scales, 
C  and  D,  just  alike  (Z>  on  the  rule  and  C  on  the  slide). 

The  student  will  note  that  the  distance  from  1  to  2  on 
scales  A  and  B  is  the  same  as  the  distance  from  2  to  4  and 


451 


from  4  to  8.  This  means  that  if  we  add  the  distance  from 
1  to  2  on  scale  B  to  the  distance  from  1  to  2  on  scale  A  by 
means  of  the  slide  B,  we  shall  obtain  the  product  2x2, 
or  4.  In  like  manner,  if  we  add  the  distance  from  1  to 
4  on  scale  B  to  the  distance  from  1  to  2  on  scale  A,  we 
shall  obtain  the  product  4  x  2,  or  8. 

C  and  D  differ  from  A  and  B  in  being  graduated  to  a 
unit  twice  as  large  as  the  unit  to  which  A  and  B  are 
graduated,  so  that  the  length  representing  the  logarithm  of 
a  given  number  on  C  and  D  is  twice  as  long  as  the  length 


A  1 


D 


FIG.  303 

on  A  and  B  representing  the.  logarithm  of  the  same  number. 
Therefore,  any  number  on  the  lower  rule  or  slide  is  oppo- 
site its  square  on  the  upper  rule  or  slide ;  and  if  the  upper 
rule  and  slide  be  regarded  as  standard  logarithmic  scales, 
the  lower  rule  and  slide  will  be  standard  logarithmic  scales 
of  the  squares  of  the  numbers  shown.  The  student  can 
easily  understand  from  the  preceding  statement  how  the 
square  roots  of  numbers  are  found. 

It  should  be  said,  however,  that  the  values  of  the  loga- 
rithms themselves  are  not  shown  on  the  scales.  What  we 
find  is  the  numbers  which  correspond  to  the  logarithms. 
Each  unit  length  on  the  scales  (graduated  lengths)  repre- 
sents equal  parts  of  the  logarithmic  table.  Thus,  if  the 
logarithm  of  10  be  selected  as  the  unit,  then  the  logarithm 


452 


of  3,  or  0.477,  will  be  represented  by  0.477  of  that  unit; 
4  by  0.602 ;  5  by  0.699 ;  and  so  on,  as  can  be  seen  by 
referring  to  the  following  table  of  corresponding  values : 


Number 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

Logarithm 

0 

0.301 

0.477 

0.602 

0.699 

0.778 

0.845 

0.903 

0.954 

1.000 

The  numbers  between  1  and  2,  2  and  3,  3  and  4,  and 
so  on,  are  represented  on  the  logarithmic  scales  by  inter- 
mediate divisions,  and  the  entire  scale  has  been  graduated 
as  closely  as  is  possible  for  convenience  in  reading. 

The  preceding  discussion  should  therefore  make  it  clear 
that  at  the  ^y^th  division  along  the  scale  on  the  slide  rule 
we  should  find  2  and  not  its  logarithm,  and  at  the  ^nnj^h 
division  we  should  find  5  and  not  0.699,  and  so  on. 

It  is  clear  that  this  scheme  eliminates  entirely  the 
process  of  finding  the  numbers  corresponding  to  certain 
logarithms,  as  we  had  to  do  when  we  computed  by  means 
of  the  table  in  Art.  492. 

The  student  will  observe,  further,  that  the  left  index  of 
the  scales  (that  is,  the  division  marked  1)  may  assume 
any  value  which  is  a  multiple  or  a  decimal  part  of  1  (for 
example,  10,  100,  1000,  0.1,  0.01,  0.001,  etc.),  but  when 
these  values  are  assumed,  this  same  ratio  must  be  held 
throughout  the  entire  scale.  In  this  case  the  proper  values 
for  the  subsequent  divisions  of  the  scale  in  order  would  be 
20,  30  ;  200,  300  ;  2000,  3000  ;  0.2,  0.3  ;  0.02,  0.03  ;  0.002, 
0.003.  It  follows  that  as  the  value  of  the  1  at  the  begin- 
ning of  each  scale  varies  any  number  such  as  382  may 
have  the  value  38200,  3820,  382,  38.2,  3.82,  0.382,  0.0382, 
and  so  on. 

If  a  number  which  the  student  has  to  read  does  not 
come  exactly  at  a  graduation  he  must  estimate  the  values 


THE  SLIDE  RULE  453 

as  closely  as  possible  ;  for  example,  if  a  certain  number  were 
indicated  ^  of  the  way  from  152  to  153,  he  would  read 
152.3,  on  the  assumption,  of  course,  that  the  left  index  of 
the  scale  has  the  value  100.  We  shall  now  proceed  to 
show  by  specific  examples  how  the  slide  rule  is  used. 

510.  Multiplication  with  the  slide  rule.  All  calculations 
in  multiplication,  division,  and  proportion  are  worked  out 
on  scales  C  and  Z>,  as  by  reason  of  the  greater  space  allotted 
to  each  of  the  divisions  the  results  obtained  are  more 
accurate.  To  find  the  product  2x3  set  scale  C  so  that 
its  left-hand  index  (the  division  marked  1)  falls  exactly 
opposite  the  division  marked  2  on  scale  D  (see  Fig.  302). 
Then  directly  opposite  the  division  marked  3  on  scale  C 
we  shall  find  on  scale  D  the  division  marked  6,  which  is 
the  required  product. 

This  process  is  justified  by  the  fact  that  to  log  2  on 
scale  D  we  add  log  3  on  scale  (7,  thus  obtaining  log  6 
on  scale  D. 

In  general,  to  multiply  any  constant  number  a  by  another 
number  ft,  set  1  of  scale  C  opposite  a  of  scale  D  and  read  the 
product  ab  on  scale  D  opposite  b  of  scale  C. 

EXERCISES 

1.  Use  the  slide  rule  to  find  the  products  in  the  following 
problems : 

(a)  2  and  4 ;  2  and  5 ;  2  and  6 ;  2  and  7  ;  2  and  8 ;  2  and  9  ; 

2  and  10. 

(b)  3  and  2 ;  3  and  3 ;  3  and  4 ;  3  and  5 ;  3  and  6 ;  3  and  7  ; 

3  and  8 ;  3  and  9 ;  3  and  10. 

2.  How  would  you  find  the  product  of  20  and  30  ? 

NOTE.  The  Mannheim  Slide  Rule  will  enable  us  to  secure  results 
correct  to  three  significant  figures,  and  in  exceptional  cases  results 


454  GENERAL  MATHEMATICS 

correct  to  even  four  significant  figures  may  be  obtained.  However, 
in  the  work  of  this  chapter  we  shall  be  content  if  we  make  our  com- 
putation correct  to  two,  or  perhaps  three,  significant  figures,  because 
in  actual  practice  this  is  sufficient. 

511.  Division  with  the  slide  rule.    To  divide  6  by  3,  set 
3  of  scale  C  opposite  6  of  scale  D  (see  Fig.  302)  and  read 
the  quotient  2  on  scale  D  opposite   1   of  scale  C.     This 
process  is  justified  from  the  fact  that  from  log  6  we  sub- 
tract log  3,  thus  obtaining  log  2  on  scale  D. 

In  general,  to  divide  any  constant  number  a  by  another 
number  6,  set  b  of  scale  C  opposite  a  of  scale  D  and  read  the 

quotient  -  on  scale  D  opposite  1  of  scale  C. 

EXERCISE 

1.  Use  the  slide  rule  to  find  the  quotients  in  the  following 
problems : 

(a\    4.6-8.10        /Vl    8  .    8  •     8  .    8        (  n\    10-     10.    10-    10 
W    2'    2'    2>    ~2~"      W   f »   f '    *»    f*      \-    )    "3"'    ~4~'    ~§~'    TO* 

512.  The  runner.     Each  slide  rule  is  equipped  with  a 
runner  which  slides  along  the  rule  in  a  groove  and  by 
means  of  which  the  student  is  enabled  to  find  more  quickly 
coincident  points  on  the  scales.    It  is  also  valuable  to  mark 
the  result  of  some  part  of  a  problem  which  contains  several 
computations.    Thus,  if 

25.2  x  3.5  x  3.68 
-22*- 

we  can  compute  25.2  x  3.5  by  one  setting  of  the  slide  and 
then  bring  the  runner  to  88.2,  the  result.  We  then  bring 
the  index  of  the  slide  to  88.2  and  multiply  by  3.68 ;  this 
gives  324.6  (approx.).  Bring  the  runner  to  this  result 
and  divide  by  22.6;  the  quotient  should  be  14.4  (approx.). 


THE  SLIDE  RULE  455 

The  student  can  easily  determine  the  use  of  the  runner 
in  finding  powers  and  roots  after  he  has  read  the  next 
two  articles. 

513.  Raising  to  powers  with  the  slide  rule.    The  student 
will  observe  that  if  the  logarithm  of  any  number  in  the 
table  of  Art.  492  be  multiplied  by  2,  the  logarithm  thus 
obtained  will  correspond  to  the  square  of  that  number. 
Thus,  if  the  logarithm  of  2  (0.301)  be  multiplied  by  2,  the 
result   (0.602)  is  the   logarithm   of   22,  or  4.    This  is  in 
accord  with  the  law  of  Art.  494  regarding  the  logarithm 
of  a  number  raised  to  a  power. 

In  like  manner,  the  logarithm  of  2  multiplied  by  3  is  0.903, 
which  is  the  logarithm  of  23,  or  8,  and  so  on.  Since  this  same 
relation  holds  for  any  number  to  any  power,  we  may  raise 
a  number  to  any  power  by  using  the  slide  rule  as  follows : 

1.  Squares  of  numbers. 

To  find  the  value  of  S2  look  for  3  on  scale  D  and  read 
3?  =  9  directly  opposite  3  on  scale  A. 

2.  Cubes  of  numbers. 

To  find  the  value  of  3s,  set  1  of  scale  B  opposite  3  of 
scale  A  and  find  3s,  or  27,  on  scale  A  opposite  3  of  scale  C. 

3.  Fourth  power  of  numbers. 

To  find  the  value  of  3*,  set  1  of  scale  C  to  3  on  scale  D 
and  find  3*,  or  81,  on  scale  A  opposite  3  of  scale  C. 

4.  Higher  powers. 

Higher  powers  of  numbers  may  be  found  by  a  method 
similar  to  the  preceding,  but  we  shall  not  need  to  discuss 
these  here,  as  most  of  our  problems  will  deal  with  the  lower 
process  of  numbers. 

514.  Square  roots  found  by  means  of  the  slide  rule.    To 
find  the  square  root  of  any  number,  bring  the  runner  to 
the  number  on  scale  A,  and  its  square  root  will  be  found 


456  GENERAL  MATHEMATICS 

at  the  runner  on  scale  D  exactly  opposite  the  number. 
This  process  is  seen  to  be  the  exact  inverse  of  finding  the 
square  of  a  number. 

If  the  number  contains  an  odd  number  of  digits  to  the 
left  of  the  decimal  point,  its  square  root  will  be  found  on 
the  left  half  of  the  rule ;  if  it  contains  an  even  number 
its  square  root  will  be  found  on  the  right  half.  If  the 
number  is  a  fraction,  and  contains  an  odd  number  of  zeros 
to  the  right  of  the  decimal  point,  the  root  is  on  the  left 
half ;  if  it  contains  an  even  number  (or  no  zeros)  the  root 
is  on  the  right  half.  If  the  student  prefers  he  may  deter- 
mine the  first  figure  of  the  root  mentally  and  then  find 
the  proper  half  of  the  rule  to  use  by  inspection. 

EXERCISES 

1.  Find  the  square  roots  of  the  following  numbers  by  means 
of  the  slide  rule  and  compare  your  results  with  those  of  the 
table  in  Art.  449  :  169 ;  576 ;  625 ;  900 ;  2.25 ;  3.24 ;  1.96 ;  4.41. 

2.  What  is  the  side  of  a  square  whose  area  is  784  sq.  ft.? 

515.  Cube  roots  found  by  means  of  the  slide  rule.  To 
find  the  cube  root  of  a  number  by  means  of  the  slide  rule, 
move  the  slide  either  from  left  to  right  or  from  right  to 
left  until  the  same  number  appears  on  scale  B  opposite  the 
given  number  on  scale  A  as  appears  on  scale  D  opposite  the 
left  or  right  index  (division  1)  on  C.  The  number  which 
appears  on  both  scales  B  and  D  is  the  cube  root  of  the 
given  number.  For  example,  to  find  the  cube  root  of  125, 
move  the  slide  to  the  left  till  5  on  scale  B  appears  opposite 
125  of  scale  A,  and  5  on  scale  D  lies  opposite  1  (the  right 
index)  on  scale  C.  Thus  5  is  the  cube  root  of  125. 

A  second  method  of  finding  the  cube  root  is  to  invert 
the  slide  (see  Art.  517)  and  set  1  on  scale  C  under  the 


THE  SLIDE  RULE  457 

given  number  on  scale  A  and  then  find  the  number  on 
scale  B  which  lies  opposite  the  same  number  on  scale  D. 
This  number  will  be  the  cube  root  sought.  The  student 
should  observe  that  the  process  of  extracting  the  cube 
root  is  the  inverse  of  that  of  cubing  a  number. 

EXERCISES 

1.  Find  the  cube  roots  of  the  following  numbers  by  means 
of  the  slide  rule  and  compare  your  results  with  those  of  the 
table  in  Art.  449  :  64  ;  125  ;  0.729 ;  2197  ;  0.001331. 

2.  Find  one  edge  of  a  cube  whose  volume  is  1728  cu.  in. 

516.  Proportion.    A  great  many  problems  in  proportion 
can  be    solved   very   easily  by  means    of  the   slide  rule. 
The  student  will    observe  that  if   the    left-hand  indexes 
of  scales   C  and  D  coincide,   the    readings    on   the   slide 
are  in  direct  proportion  1:1  to  the  opposite  readings  on 
the  rule. 

In  like  manner,  if  the  left  index  of  C  is  made  to  coin- 
cide with  2  on  D,  the  ratio  will  be  1 :  2,  and  so  on.    Hence, 

10      25 

to   find   the  fourth  term   in   the  proportion  —  =  — ,  we 

15       x 

move  the  slide  to  the  right  until  10  on  scale  C  is  opposite 
15  on  scale  D,  and  opposite  25  on  C  read  off  x  =  37.5  on  D. 
In  general,  to  find  the  fourth  term  in  a  proportion,  set  the 
first  term  of  the  proportion  on  scale  C  to  the  second  term  on 
scale  D  and  find  the  fourth  term  (unknown)  on  scale  D  oppo- 
site the  third  term  on  C. 

517.  Inverted  slide.    The  slide  may  be  inverted  so  that 
scale  C  faces  scale  A.     This  gives  inverted   readings  on 
scale  C  which  are  the  reciprocals  of  their  coincident  read- 
ings on  scale  D,  and  vice  versa.    Thus,  3  on  scale  D  lies 
opposite  0.33  on  scale  (7,  and  vice  versa. 


458  GENERAL  MATHEMATICS 

The  inverted  slide  is  useful,  as  is  shown  in  finding  cube 
roots  and  also  in  problems  involving  an  inverse  propor- 
tion, as  in  the  following  example : 

If  10  pipes  can  empty  a  cistern  in  22  hr.,  how  long  will  it 
take  50  pipes  to  empty  the  cistern  ? 

Solution.  Invert  the  slide  and  set  10  of  scale  B  at  22  on  scale  D, 
and  opposite  5  on  C  find  4.4  (the  result)  on  scale  D. 

518.  Position  of  the  decimal  point.  The  student  will  be 
able  in  most  practical  problems  to  determine  the  position 
of  the  decimal  point.  If  there  is  any  considerable  difficulty 
in  any  later  work  he  should  consult  some  standard  manual 
on  the  use  of  the  slide  rule. 

MISCELLANEOUS  EXERCISES 
Solve  the  following  problems  by  means  of  the  slide  rule : 

1.  Find  the  product  of  58.2  x  2.55;  33.4  x  75.6;  22.5  x 
33.3  x  8.2 ;  0.12  x  0.09  x  0.003. 

82.5        3.5        0.04 

2.  tind    the     following    quotients:     — — ;     — — ;     ;— ; 

35.3  x  75.5 .    7.2  x  83.5  x  0.09 
22.8  3.6 

3.  Perform  the  indicated  operations:  252;  3  •  33;  2  x  22  x  158; 
124 ;  1.22  x  7.52  x  0.92. 

4.  Extract  the  indicated  roots:   V2?  V3;  V5;  V7;  Vl2; 
V576;  VTO6;  Vl37.2 ;    -^8;   -^15;   -\/Tl2;   A/64;   ^1728. 

5.  Find  the  circumference  of  a  circle  whose  diameter  is  6; 
5.5 ;  2.83.    (See  a  slide-rule  manual  for  a  short  cut.) 

6.  If  10  men  can  do  a  piece  of  work  in  4  da.,  how  long 
will    it   take    5   men  to  do  the  work   if   they    work   at   the 
same  rate? 

7.  What  will  be  the  cost  of  13|  ft.  of  rope  at  3j$  per  foot  ? 


THE  SLIDE  RULE  459 

8.  What  is  the  volume  of  a  cubical  stone  6.3  ft.  long,  4.5  ft. 
wide,  and  3.2  ft.  high  ? 

9.  What  distance  will  a  train  travel  in  10  hr.  and  20  min. 
at  a  rate  of  30.5  mi.  per  hour  ? 

10.  What   is   the    interest  on    $5600  for  1  yr.    at   3J%  ? 

at  5%?  at  6%? 

11.  In  how  many  hours  will  a  train  travel  756  mi/ if  it 
travels  at  the  rate  of  41.2  mi.  per  hour  ? 

12.  How  many  miles  in  2783  ft.  ?  in  17,822  ft.  ? 

13.  Find  the  area  of  a  circle  whose   diameter   is   10  in.; 
7.5  in.;  0.351  in.    (Use  A  =  0.7854 d2.) 

14.  Find   the  volume    of   a  cube  one  of  whose   edges    is 
3.57  in. ;  of  a  cube  one  of  whose  edges  is  82.1  in. 

15.  If  a  man  can  save  $3120  in  26  mo.,  how  many  dollars 
will  he  save  at  the  same  rate  in  1  mo.  ?  6  mo.  ?  12  mq.  ? 

16.  Find  the  mean  proportional  between  6  and  27. 

17.  If  goods  cost  550  a  yard,  at  what  price  must  they  be 
sold  to  realize  15%  profit  on  the  selling  price  ? 

18.  The  formula  for  the  area  of  the  ring  in  Fig.  304  is 
A  =  irR*  —  Trr2.    Since  2  r  =  d,  this  formula  may  be 

written  A-        £=-  1,   or  A=        ±~^J2~ 

Using  the  last  formula,  find  the  area  of  the  ring 

when  R  =  8.5  in.  and  r  ='  5.3  in.  FIG.  304 

19.  Find  the  amount  of  $225  invested  for  12  yr.  at  6% 
simple  interest.    (Use  the  formula  found  under  logarithms  or 
refer  to  a  slide-rule  manual  for  a  short  cut.) 

20.  What  force  must  be  applied  to  a  lever  5.2  ft.  from  the 
fulcrum  to  raise  a  weight  of  742  Ib.  which  is  1|  ft.  on  the 
opposite  side  of  the  fulcrum  ? 


460 


GEXK 1 1 A  L  M  AT  H  KM  ATICS 


21.  A  shaft  makes  28  revolutions  and  is  to  drive  another  shaft 
which  should  make  42  revolutions.    The  distance  between  their 
centers  is  60  in.    What  should  be  the 

diameter  of  the  gears  ?  (See  Fig.  305.) 

HINT.  Refer  to  a  slide-rule  manual 
or,  better,  develop  the  formula  between 
the  diameter  of  the  gears,  the  number 
of  revolutions,  and  the  distance  between 
their  centers,  as  follows  :  FIG.  305 

28  circumferences  of  the  large  gear  =  42  circumferences  of  the  small 
gear.  Why '.' 

Then  28  (60  -  x)  =  42  x  (see  Fig.  305). 

28  •  60  =  70  x.  Why  ? 

|  =  f. 

Then  apply  the  slide  rule. 

22.  Since  F  =  ^C  +  32  is    the  equation   representing  the 
relation  between  the  Fahrenheit  and  centigrade  thermometers, 
we  may  compare  readings  of  these  two  thermometers  by  the 
following  scheme : 


c 

Set  5 

Below  degrees  centigrade 

I) 

to  9 

Read  degrees  plus  32  equals  Fahrenheit 

Wrhat  then  is  F.  when  C.  =  25°  ?  18°  ? 

*23.  Trigonometric  applications  are  greatly  simplified  by  the 
slide  rule.   The  solution  of  a  formula 

be  sin^4  .  . 

like  A  =  — is  readily  obtained. 

Sin  A  may  be  used  directly  as  a  factor 
in  performing  the  operation.  Find  the 
area  of  the  corner  lot  in  Fig.  306. 


6=84' 

FIG.  306 

*24.  Find  the  value  of  the  lot  in  Ex.  23  at  $871.20  per  acre. 

NOTE.    For  numerous  applications  of  the  slide  rule  to  practical 
problems  the  student  should  consult  a  standard  slide-rule  manual. 


THE  SLIDE  KULE  461 

SUMMARY 

519.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words  and  phrases:  slide  rule,  runner,  inverted  slide. 

520.  The  theory  and  practical  value  of  the  slide  rule 
have  been  discussed  and  illustrated  so  that  the  student 
can  get  at  least  an  elementary  working  knowledge  of  this 
powerful  labor-saving  device. 

521.  The  student  has  been  taught  how  to  use  the  slide 
rule  in  multiplying,  dividing,  raising  to  powers,  and  ex- 
tracting roots. 

522.  The    student   has  been  shown   how   problems    in 
proportion  and  many  other  verbal  problems  may  be  solved 
by  the  slide  rule. 

523.  The  student  has  been   referred  to  the   slide-rule 
manual  for  methods  of  solving  the  more  difficult  problems. 


CHAPTER  XIX 

QUADRATICS;  QUADRATIC  FUNCTIONS;  QUADRATIC  EQUA- 
TIONS;  GRAPHS  OF  QUADRATIC  EQUATIONS;  FORMULAS 
INVOLVING  QUADRATIC  TERMS 

524.  Quadratic-equation  problem.  An  engineer  increased 
the  speed  of  his  train  2  mi.  an  hour  and  made  a  run  of 
180  mi.  in  1  hr.  less  than  schedule  time.  What  was  the 
speed  when  running  according  to  schedule  ? 

Solution.    Let     x  =  the  ordinary  rate  of  the  train. 
Then  x  +  2  =  the  rate  after  the  increase. 

180 

— —  =  the  schedule  time. 
x 

180 

=  the  time  it  takes  after  the  speed  is  increased. 


x  +  2 

180       180 

Then  —  =  —    —  hi.  A\  hv  .; 

x        x  +  2 

The  L.C.D.  is  x  (x  +  2).  Why  ? 

Multiplying  through  by  x  (x  +  2)  we  get 
180  (x  +  2)  =  ISOx  +  x(x  +  2). 
180  x  +  360  =  180  x  +  x2  +  2  x. 

360  =  x*  +  2x.  Why  ? 

x2  +  2x-  360  =  0.  Why? 


We  are  not  able  to  simplify  the  equation  o?+  2#—  360  =  0 
further.  The  methods  of  the  preceding  chapters  will  not 
solve  it.  In  fact  we  appear  to  have  come  to  the  end  of 
the  road.  It  is  clear  that  the  problem  is  solved  if  we 
can  find  a  value  of  x  which  will  make  the  value  of  the 
quadratic  trinomial  z2  +  2  x  —  360  equal  to  zero. 

462 


QUADRATIC  EQUATIONS 


463 


An  equation  in  which  the  highest  power  of  the  unknown 
is  the  second  power  is  called  a  quadratic  equation.  Many 
problems  in  geometry,  science,  and  mechanics  are  solved 
by  quadratic  equations.  It  is  our  purpose  in  this  chapter 
to  develop  the  power  to  solve  quadratic  equations  and  to 
apply  quadratic  methods  to  verbal  problems.  This  process 
will  be  illustrated  by  the  solution  of  the  given  equation, 
3?  +  2  x  -  360  =  0. 

525.  Quadratic  function.    The  expression  x*  +  2  x  —  360 
is  a  quadratic  function  of  x,  or  a  function  of  the  second 
degree ;  with  every  change  in  the  value  of  x  the  value  of 
the  function  x2  +  2  x  —  360  changes.    We  shall  get  some 
light  on  the  question,  What  value  of  x  will  make  the  ex- 
pression 3?  +  2x  —  360  equal  to  0  ?  by  studying  how  the 
value  of  the  expression  x2  +  2  x  —  360  changes  as  we  give 
different  values  to  x.    This  variation  is   best  shown  by 
means  of  the  graph. 

526.  Graph  of  a  quadratic  function.    In  order  to  under- 
stand more  about  the  graph  of  a  quadratic  function  we 
shall  consider  a  few  simple  exercises. 

INTRODUCTORY  EXERCISES 

1.  Find  the  value  of  the  function  a-2  +  2  a-  —  360  for  each  of 
the  following  values  of  x :  0, 10,  — 10,  - 15,  15,  20,  18, 19,  21. 

2.  Fill    in    the    following    table   of    corresponding   values 
for  x  and  the  function  x2  +  2  x  —  360 : 


X 

0 

10 

-10 

+  15 

-15 

+  16 

-16 

+  17 

-17 

J2  +  2x-360 

-360 

-240 

-280 

x 

+  19 

-19 

+  21 

-21 

+  25 

-25 

+  30 

-30 

x2  +  2x-360 

464 


GENERAL  MATHEMATICS 


If  we  transfer  the  corresponding  values  of  x  and  the  function 
x2  +  2  x  —  360  from  the  table  to  a  sheet  of  cross-section  paper 


f(t 

\ 

1 

1 

\ 

I 

f 

V 

i 

\ 

1 

f 

v 

i 

\ 

1 

\ 

\ 

1 

f 

v 

1 

^ 

f 

v 

1 

*y 

( 

) 

?,< 

\ 

-i 

(} 

(i 

1 

(1 

I 

? 

) 

?, 

n 

' 

f 

\ 

f 

^ 

f 

V 

J 

s 

/ 

s 

f 

s 

, 

/ 

s 

^ 

^ 

FIG.  307.   SHOWING  THE  GRAPH  OF  A  QUADRATIC  FUNCTION 

so  as  to  secure  the  points  which  correspond  to  these  values, 
we  shall  obtain  a  series  of  points  which,  when  connected  by 
a  smooth  curve,  will  be  like  the  curve  in  Fig.  307. 

EXERCISES 

1.  From  the  graph  (Fig.  307)  determine  how  the  value  of 
a-2  _j_  2  x  _  360  changes  as  x  changes  from  25  to  20 ;  from  0  to 
15 ;  from  0  to  -  15 ;  from  -  25  to  -  30 ;  from  +  25  to  +  30. 

2.  What  value  of  x  will  make  the  function  or2  +  2x  —  360 
equal  to  300?  200?  150?  -250?  -300? 

The  preceding  exercises  show  us  that  the  graph  enables 
us  to  see  what  value  x  must  have  in  order  that  the  expres- 
sion a^+  2 x—  360  shall  have  a  given  value.  The  pupil  will 
recall  that,  in  the  speed  problem  with  which  we  started,  the 


QUADRATIC  EQUATIONS  465 

one  question  we  could  not  answer  was,  What  value  of  x 
will  make  the  expression  a^+  2  x  —  360  equal  to  zero  ?  This 
question  is  now  easily  answered.  A  glance  at  the  graph 
shows  that  the  expression  becomes  0  at  two  places  ;  namely, 
when  a:  =  18  or  when  a;  =  —  20.  Thus  the  equation  a?+2x 
—  360  =  0  is  satisfied  by  x  =  18  or  x  =  —  20.  Check  these 
values  by  substituting  in  the  equation.  Hence  the  speed 
of  the  train  running  according  to  schedule  was  18  mi.  per 
hour.  We  reject  the  —  20  as  meaningless.  (Why  ?) 

527.  The  two  solutions  of  a  quadratic  equation.    In  the 
preceding  article  we  rejected  —  20  as  a  meaningless  solution. 
Though  a  quadratic  equation  has  two  solutions,  this  does 
not  mean  that  every  verbal  problem  that  leads  to  a  quad- 
ratic equation  has  two  solutions.    The  nature  of  the  con- 
ditions of  the  verbal  problem  may  be  such  as  to  make  one 
or  even  both  of  the  solutions  of  the  quadratic  impossible 
or  meaningless.    When  neither  of  the  two  solutions  of  the 
quadratic  is  a  solution  of  the  problem  it  usually  means 
that  the  conditions  of  the  problem  are  impossible,  or  that 
the  problem  is  erroneously  stated.    In  fact  it  would  be 
easy   to   make   up   an    arithmetic    problem   whose   result 
could  not  be  interpreted ;  for  example,  it  would  be  diffi- 
cult to  jump   out   of  the  window  2i  times.    To  decide 
which  solution,  if  either,  meets  the  conditions  stated  in 
a  verbal  problem,  it  is  necessary  to  reread  the  problem, 
substituting  the  solutions  in  the  conditions  of  the  prob- 
lem,  and   to  reject  solutions   of   the  equation  which  do 
not  meet  the  conditions. 

528.  How  to  solve  a  quadratic  equation  graphically.    We 
may  now  summarize  the  method  of  solving  a  quadratic 
equation  revealed  in  the  discussion  of  Arts.  526-527  as  fol- 
lows :  (1)  Reduce  the  equation  to  the  form  ax2  +  bx  +  c  =  0. 
(2)  Make  a  table  showing  the  corresponding  values  of  x  and 


460 


G EN  ER  A  L  M  AT  1 1 E  M  ATICS 


the  function  ax2  -f  bx  +  c.  (3)  Transfer  the  data  of  the 
table  to  squared  paper  and  construct  the  curve  represent- 
ing the  function  ax2  +  bx  +  c.  This  curve  shows  the  values 
of  ax2  +  bx  +  c  which  correspond  to  the  different  values 
of  x.  (4)  By  inspection  determine  the  points  of  the  curve 
where  the  expression  is  zero.  The  values  of  x  for  these 
points  are  the  solutions  of  the  equation. 

EXERCISES 

Solve  the  following  equations  graphically,  and  check  : 
1.  ic2-  9z-f  14  =  0. 

Plot  the  function  between  the 
limits  0  to  12.  This  means  con- 
struct the  table  by  letting  x  equal 
the  following  values :  0,  1,  2,  3, 
4  ...  12. 


6.  4a2 

Plot  from  —  2  to  5. 

Plot  from  -  4  to  +  2. 

8.  100  x2-  5  x  -495  =  0. 
Plot  from  —  5  to  4-  5. 

9.  6a;2-17x  =  20. 
Subtract  20  from  both  mem- 


2. z2-6z  +  5  =  0. 
Plot  from  - 1  to  7. 

3.  x2-  3x  -10  =  0. 
Plot  from  -  3  to  6. 

4.  x2  -11  x  +  24  =  0. 
Plot  from  1  to  10. 

5.  x*-llx  +  25  =  0. 
Plot  from  1  to  10. 

11.  9x2  +  3x  +  20 

Subtract  2  x2  +  2  x  +  50  from  both  members  of  the  equation. 

529.  The  graph  solves  a  family  of  equations.  At  this 
point  the  student  should  note  that  a  single  graph  may  be 
used  to  solve  a  whole  family  of  equations.  Thus,  if  we  turn 
to  Fig.  307  we  see  that  the  curve  for  x2—  2x—  360  can 


bers  and  plot  6  x2  -11  x-  20. 

10.  2x2-  9  =  3x. 

Subtract  3  x  from  both  mem- 
bers and  plot  the  function 
2:r2-9  -3z. 


QUADBATIC  EQUATIONS  467 

be  used  not  only  to  solve  the  equation  x*  +  2  x  —  360  =  0  but 
also  to  solve  every  equation  of  the  type  a?  +  2  x  —  360  =  c 
(where  c  is  some  arithmetical  number).  For  example,  if  we 
ask  what  value  of  x  will  make  xz  +  2  x  —  360  equal  to  100, 
we  can  tell  by  looking  at  the  curve  that  the  answer  is  20.5 
or  —  22.5,  and  this  is  precisely  the  same  as  saying  that  the 
two  roots  of  the  equation  j?  +  2  x  —  360  =  100  are  20.5 
and  -  22.5. 

EXERCISES 

Solve  by  the  graph  : 

1.  x2  +  2x-  360  =  200.  6.  x2  +  2  x  =  400. 

2.  **  +  2*  -  360  =180. 


Subtract  36Q 

3.  x-  +  2  x-  360  =  400. 

4.  &  +  2x  -  360  =  -  250.  ?•  x2  +  2x  -  500  =  0. 

5.  x2  +  2x-  360  =  -360.  Add  140.   Why? 

The  last  two  exercises  show  that  the  graph  solves  all 
equations  in  which  x*  +  2  x=  c  (some  arithmetical  number). 
For  we  can  write  the  given  equation  x2  +  2  x  —  500  =  0  in 
the  form  x*  +  2  x  —  360  =  140.  This  last  form  we  are  able 
to  solve  at  sight  by  the  graph. 

530.  The  parabola.  The  curve  representing  the  func- 
tion x*+  2x—  360  shown  in  Fig.  307  is  called  a  parabola. 
Study  and  discuss  the  general  shape  and  symmetry  of  the 
curve.  Compare  the  curves  you  and  your  classmates  have 
drawn  in  the  exercises  of  this  chapter  and  see  if  you  can 
find  a  parabola  in  an  earlier  chapter  of  this  text. 

The  graph  of  a  quadratic  function  in  one  unknown  is 
a  parabola.  iWis  a  symmetrical  curve.  No  three  points  of 
the  curve  lie  on  a  straight  line.  The  parabola  is  a  common 
notion  in  physics  and  mechanics.  Thus,  the  path  of  a 
projectile  (for  example,  a  bullet)  is  a  parabola.  A  knowl- 
edge of  the  theory  and  application  of  many  such  curves 


468  GENERAL  MATHEMATICS 

was  of  extreme  importance  in  the  recent  world  war.  The 
soldiers  who  had  been  trained  in  some  of  the  more  advanced 
mathematical  courses,  especially  in  trigonometry  and  graphi- 
cal work,  were  in  demand  and  were  given  plenty  of  oppor- 
tunity to  put  into  practice  what  they  had  learned  in  school. 

In  plotting  functions  like  x?  +  2  x  —  360  we  plot  the 
values  of  x  along  the  a>axis  and  the  corresponding  values 
of  the  function  on  the  ?/-axis.  This  suggests  that  the 
curve  obtained  in  Fig.  307  is  the  graph  of  the  equation 
y  =  z2  +  2  a:  -  360. 

It  follows  that  whenever  y  is  a  quadratic  function  of  x, 
or  when  x  is  a,  quadratic  function  of  y,  the  graph  of  the 
equation  is  a  parabola. 

EXERCISES 

Graph  each  of  the  following  equations : 

1.  y  =  a;2-4.  3.  z  =  y2  +  5  ?/ +  4. 

2.  y  =  x2  +  3  x  +  2.  4.  x  =  f  -  1  y  +  6. 

531.  Maxima  and  minima.  The  theory  of  maxima 
(greatest  values)  and  minima  (least  values)  of  functions 
has  many  important  applications  in  geometry,  physics,  and 
mechanics. 

This  article  will  present  one  example  drawn  from  each 
subject.  A  careful  study  of  the  following  example  will 
suggest  the  proper  method  of  attack. 

ILLUSTRATIVE  EXAMPLE 

A  rectangular  garden  is  to  be  inclosed  on  ^iree  sides,  the 
fourth  side  being  bounded  by  a  high  wall.  WlJft  is  the  largest 
garden  that  can  be  inclosed  with  20  rd.  of  fencing  ? 

Solution.  Let  x  represent  the  width. 

Then  20  —  2  x  represents  the  length, 

and  20  x  —  2  x2  represents  the  area. 


469 


We  are  now  interested  in  a  maximum  (greatest  possible)  value 
that  20  x  —  2  x2  can  have.  By  trial  we  obtain  the  corresponding 
values  for  x  and  the  function  20  x  —  2  x2  shown 
in  the  table  below. 

Common  sense,  the  table,  and  the  curve  of 
Fig.  308  show  us  that  if  the  garden  is  made 
very  wide  or  very  narrow  the  area  is  very  small. 
The  table  and  the  curve  of  Fig.  308  suggest  that 
50  is  probably  the  largest  area.  In  this  case  the 
dimensions  of  the  garden  are  5  and  10.  By 
taking  x  first  a  little  larger  and  then  a  little 
smaller  than  5,  we  may  check  our  conclusion. 


We  can  as  a  matter  of  fact,  save  our- 


X 

20z  —  2x2 

0 

0 

1 

18 

2 

32   . 

3 

42 

4 

48 

5 

50 

6 

48 

7 

42 

10 

0 

selves  much  of  the  labor  of  these  computations  by  an  alge- 
braic method,  which  we  shall  present  in  Art.  538.    At  this 


-/(*> 


-20 


FIG.  308.  SHOWING  THE  MAXIMUM  VALUE  or  A  QUADRATIC  FUNCTION 

stage,  however,  we  shall  be  content  to  plot  the  curves  and 
find  the  highest  or  the  lowest  point  on  the  curve. 


470  GENERAL  MATHEMATICS 

EXERCISES 

1.  If  a  ball  is  thrown  upward  with  a  velocity  v0,  the  dis- 
tance d  from  the  earth  to  the  ball  after  a  given  time  t  is  given 
by  the  physics  formula 

d  =  v0t-  16 12. 

How  high  will  a  ball  rise  which  is  thrown  with  an  initial 
velocity  of  100  ft.  per  second  ? 

HINT.  The  formula  becomes  d  =  100 1  —  16  <2.  Plot  the  function 
and  find  by  inspection  its  greatest  value. 

2.  Divide  10  into  two  parts  such  that  their  squares  shall  be 
a  minimum. 

*3.  Find  the  most  advantageous  length  of  a  lever  for  lifting 
a  weight  of  100  Ib.  if  the  distance  of  the  weight  from  the 
fulcrum  is  2  ft.  and  the  lever  weighs  4  Ib.  to  the  foot. 

532.  Limitations  of  the  graphic  method  of  solving  quad- 
ratic equations.  By  this  time  the  student  is  no  doubt  con- 
vinced that  the  graphic  method  of  solving  quadratic 
equations  has  its  limitations.  We  may  enumerate  the 
following:  (1)  The  results  are  frequently  rough  approx- 
imations. This  is  evident  the  moment  we  attack  problems 
of  some  slight  difficulty.  In  fact  the  earlier  problems  of  the 
chapter  are  artificially  built  so  that  in  all  probability  the 
student  will  accidentally  get  an  accurate  result.  We  must 
remember  that  the  graphic  method  depends  for  its  accuracy 
upon  the  mechanical  (or  nonintellectual)  conditions,  such 
as  the  skill  of  the  student  at  this  type  of  work,  the  exact- 
ness of  squared  paper,  and  our  ability  to  estimate  fractional 
parts  of  the  unit.  (2)  Aside  from  the  fact  that  the  signifi- 
cance of  a  graph  is  sometimes  obscure,  the  work  is  a  bit 
cumbersome  and  tedious.  (3)  It  is  not  economical  of  time, 
as  we  shall  presently  show. 


QUADRATIC  EQUATIONS.  471 

533.  More  powerful  methods  of  solving  quadratic  equa- 
tions.   Because  of  the  foregoing  limitations  of  the  graphic 
method   we  are  ready  to  proceed  to  the  study  of  more 
efficient  methods.    These  methods  rest  purely  on  an  intel- 
lectual   basis    (that    is,    the    accuracy    is   independent  of 
constructed  figures).    We  shall  observe  that  they  get  the 
results  quickly  and  with  absolute  accuracy. 

534.  Quadratic  equations  solved  by  factoring.   The  factor- 
ing method  may  be  illustrated  by  the  following  solution 
of  the  speed  problem  with  which  we  opened  the  chapter : 

Solution.    Given  x2  +  2  x  -  360  =  0. 
Factoring  the  left  member, 

(x  +  20)  (x  -  18)  =  0. 

The  preceding  condition,  namely,  (a;  +  20)  (z  —  18)  =  0,  will  be 
satisfied  either  if  x  +  20  =  0  or  if  x  — 18  =  0,  for  we  learned  in 
Art.  236  that  the  product  of  two  numbers  is  zero  if  either  factor 
is  zero.  Thus,  5  x  0  =  0  or  0  x  8  =  O/ 

Now  if  x  +  20  =•  0, 

then  x  -  —  20. 

And  if  x  -  18  =  0, 

then  x  - 18. 

Hence  x  -  +  20, 

or  x-- 18. 

In  the  next  solution  we  shall  omit  a  considerable  part  of  the  dis- 
cussion and  show  how  the  work  may  be  arranged  in  a  few  simple 
statements. 

Solve  z2  -  9  x  +  14  =  0. 

Factoring,  (x  -  7)  (x  -  2)  =  0.  (1) 

This  equation  is  satisfied  if    x  —  7  =  0,  (2) 

or  if  x  -  2  =  0.  (3) 

From  equation  (2)  x  =  7, 

and  from  equation  (3)  x  =  2. 

The  numbers  satisfy  the  equation,  consequently  2  and  7  are  the 
roots  of  the  equation  x'2  —  9  x  +  14  =  0. 


472  GENERAL  MATHEMATICS 

EXERCISES 

Solve  the  following  quadratic  equations  by  the  method  of  fac- 
toring and  test  the  results  by  substituting  in  the  equations  : 

1.  x2  -  5x  +  6  =  0.  15.  20x  -  51  =  x2. 

2-  if -7 y +12  =  0.  16.  77+4d  =  d2. 

4.  x2- 6x-27=0.  18.10^  +  33^=7. 

.     5.  x2-2x-35  =  0.  19.  6z2=23z  +  4. 

6.  cc2+  5x  =  6.  x      I      I 

20.  —  +  —  =  —• 
HINT.    Subtract      6     from  '2       2       x 

both  members  before   apply-  3 

ing  the  method.    Why?  21>   15a;  +  4  =  ^' 

7.  ar  +  a;  =  56.  r  _  jt       33 

o-l 

HINT.    Subtract   2x   from  oo     K  ~       Q       _. 

&O>      t-J  *)U   O    ^~~ 

both  members  and  rearrange  x 

terms  before  factoring.  y       15        ?/2 

9.  ^  +  4  =  42,.  "  2  +  T"=14' 

10.  x2-  85  =12 a-.  25.-^-+^       =4. 

11.  «2  =10^  +  24.  ~  y        0 

12.  ?/?2  —  91  =  6 w.  26-       2        =  "3       a  +  3' 

XO*      i-C      ~~ r~   3C    - — •    -r*-.  /y»  or»    

07         x      _  £ i  _  o 

14.  m2  +112  =  23m.  '  x-3      a;  +  3 

VERBAL  PROBLEMS 

Solve  the  following  problems  by  the  factoring  method  and 
test  the  results  by  substituting  the  solution  in  the  conditions 
of  the  problem  : 

1.  A  crew  rows  across  a  calm  lake  (12  mi.  long).  On  the 
return  trip  it  decreases  the  rate  by  1  mi.  per  hour  and  makes  the 
trip  in  7  hr.  Find  the  rate  of  the  crew  both  going  and  returning. 


QUADRATIC  EQUATIONS 


473 


2.  A  man  drives  'a  car  80  mi.  out  of  town.     On  the  return 
trip  he  increases  his  rate  8  mi.  per  hour.    He  makes  the  trip 
in  4^  hr.    What  was  his  rate  while  driving  out  ? 

3.  The  base  of  a  rectangle  exceeds  the  altitude  by  5  in.    The 
area  equals  150  sq.  in. 


Find    the 
altitude. 


base    and 


40 


-90-2W- 


T 

I 

40  -2W 

f 

W 

I 


— 90 — 
FIG.  309 


4.  The  base   of  a 
triangle  is  7  in.   less 
than  the  altitude.  The 
area  equals  85  sq.  in. 
Find    the    base    and 
altitude. 

5.  A  farmer  is  plowing  a  field  of  corn  40  rd.  wide  and  90  rd. 
long  (Fig.  309).    At  the  end  of  a  certain  day  he  knows  that  he 
has  plowed  five  sixths  of  the  field.    How 

wide  a  strip  has  he  plowed  around  the 
field? 

6.  A  piece   of  tin   in  the  form  of  a 
square  is  taken  to  make  an  open  box. 
The   box   is   made   by  cutting  a  1-inch 
square  from  each  corner  of  the  piece  of 
tin  and  folding  up  the  sides  (Fig.  310). 
The  box  thus  made  contains  36  cu.  in. 

Find  the  length  of  the   side   of   the   original  piece  of  tin. 

535.  Limitations  of  the  factoring  method.  In  some  of  the 
preceding  exercises  the  quadratic  expressions  were  very 
difficult  to  factor.  This  is  usually  the  case  when  the 
constant  terms  are  large  numbers.  Indeed,  most  verbal 
problems  lead  to  quadratic  equations  which  cannot  be 
solved  by  the  factoring  method.  The  following  problem  is 
a  simple  illustration: 


FIG.  310 


474  GENERAL  MATHEMATICS 

What  is  the  length  of  a  side  of  a  square  (Fig.  311)  whose 
diagonal  is  2  ft.  longer  than  a  side  ? 
Attempted  solution  by  factoring  method: 
Let  x  =  a  side  of  the  square. 

Then    x  +  2  =  the  length  of  the  diagonal,    x 
By  the  theorem  of  Pythagoras 

xz  +  x2  =  x2  +  4  x  +  4. 
Simplifying,  &  -  4  x  -  4  =  0. 

This  appears  to  be  the  end  of  the  road  ; 

we  cannot  factor  x2  —  4  x  —  4,  for  we  cannot  obtain  a  combination 
of  whole  numbers  or  fractions  whose  product  is  —  1  and  whose  sum 
is  —  4.  And  yet  we  are  probably  convinced  that  such  a  square  does 
exist  though  we  are  forced  to  admit  that  the  solution  of  the  problem 
by  the  factoring  method  is  hopeless. 

536.  Solution  of  the  quadratic  equation  by  the  method  of 
completing  the  square.  If  we  were  able  to  make  the  left 
member  of  the  equation  z2  —  4  a;  —  4  =  0  a  perfect  square 
without  introducing  the  unknown  (#)  into  the  right  mem- 
ber, we  could  take  the  square  root  of  each  member  of  the 
equation  and  thus  obtain  a  linear  equation  which  would 
be  easily  solved.  This  is  precisely  the  method  we  wish  to 
employ.  However,  we  must  first  learn  to  make  the  left 
member  x2-  —  4  x  —  4  a  perfect  square. 

ORAL  EXERCISES 

1.  Find  (x  +  2)2  ;  (x  +  3)2  ;  (x  +  4)2  ;  (a;  -  2)a. 

2.  When  is  a  trinomial  a  perfect  square  ?  (See  Art.  250.) 

3.  Make  a  perfect  square  trinomial  of  the  following:  a-2  —  6  x; 


7  x 
x2  +  7  x  ;  x2  +  9  x  ;  x2  +  —  • 


QUADRATIC  EQUATIONS  475 

The  preceding  exercises  show  that  it  is  easy  to  complete 
the  square  of  a  binomial  of  the  form  x2  +  ax,  for  we  need 
only  to  add  the  square  of  half  the  coefficient  of  x.  Then,  too, 
the  constant  term  of  a  trinomial  can  always  be  made  to 
appear  in  the  right  member  of  the  equation,  leaving  the 
left  member  in  the  form  x2  +  ax.  We  now  proceed  ,  to 
solve  the  equation  x2  —  4  #  —  4  =  0.  Write  the  equation 

/—  4\2 
thus:  Xs—  4  #  =  4.    Add  (—5-)  »  or  4,  to  make  the  first 

\  2  / 

side  a  trinomial  square,  and  we  obtain 
xz  _  4  x  _|_  4  =  8. 

Taking  the  square  root  of  both  sides  and  remembering 
that  8  has  two  square  roots,  +  V8  or  —  V8,  we  get 


(1) 
or  ar-2=-V8.  (2) 

From  equation  (1)  we  get  x  =  2  +  V8,  and  from  equation 
(2)  we  get  a:=2_-V8. 

If  we  obtain  V8  either  by  the  arithmetical  method  taught 
by  Art.  446  or  by  using  the  table  of  Art.  449,  the  result 
(accurate  to  three  places)  is  2.828.  Then  x=  2  +  2.828,  or 
4.828.  Hence  the  side  of  the  square  whose  diagonal  is  2  ft. 
longer  than  a  side  is  4.828  ft.  We  can  check  this  result  by 
applying  the  theorem  of  Pythagoras. 

We  reject  2—  V8,  or  —  0.828,  because  it  does  not  satisfy 
the  conditions  of  the  problem.  However,  the  student  should 
realize  that  —  0.828  is  just  as  much  a  solution  of  the 
equation  z2  —  4  x  +  4  =  8  as  is  4.828. 

The  method  of  completing  the  square  is  further  illus- 
trated by  the  following  solution  of  the  equation 

10  z2-  9*  +  2  =  0. 


476  GENERAL  MATHEMATICS 

Write  the  equation      10  x2  -  9  x  =  -  2.  Why '? 

9  x          1 

Dividing  by  10,  z2  -  —  =  -  -  . 

1U  5 

Note  that,  the  left  member  is  now  more  easily  completed.    Why  ? 

x»_»£ 

10 

x2_9£  +  ^L-.J_.  Why? 

10  +  400      400 

Taking  the  square  root  of  each  member, 
x  -  &  =  ±  5V 
Whence  x  =  %  or  f . 

537.  Summary  of  the  method  for  solving  quadratic  equa- 
tions by  the  method  of  completing  the  square. 

1.  /Simplify  the  equation  and  reduce  to   the  form  ax*  + 
bx  =  c. 

2.  If  the  coefficient  of  x2  is  not  1,  divide  both  members  of 
the  equation  by  the  coefficient  so  that  the  equation  takes  the 
form  x2  +px  =  q. 

3.  Find  half  the  coefficient  of  x;  square  the  result ;  add 
the  square  to  both  members  of  the  equation  obtained  in  step  2. 
This  makes  the  left  member  a  perfect  square. 

4.  Express  the  right  member  in  its  simplest  form. 

5.  Take  the  square  root  of  both  members,  writing  the  double 
sign  ±  before  the  square  root  in  the  right  member. 

6.  Set  the  left  square  root  equal  to  the  positive  root  in  the 
right  member  of  the  equation  in  5.    Solve  for  the  unknown. 
This  gives  one  root. 

7.  Repeat  the  process,  using  the  negative  root  in  5.    This 
gives  the  second  root  of  the  equation. 

8.  Express  the  roots  first  in  simplest  form. 


QUADRATIC  EQUATIONS  477 

EXERCISES 

Solve  by  the  method  of  completing  the  square,  and  check  : 

'  1.  xa-6»  =  91.  11.  4z2  +  45z-36  =  0. 

2.  x2-8a;  =  48.  12.  6z2  +  7cc-20  =  0. 

3.  X2_x_3  =  o.  13.  22  +  62=l. 

4.  y3  _|_  4y  _)_  3  =  0.  HINT.    Compute  roots  to  the 

-       o    i    A  K       n  nearest  hundredth. 

5.  if  +  4y  —  5  =  0. 

6.  /,*  _(-  8  b  -  20  =  0.  14.  x2  +  4  a  =  16. 

7.  y2  +14^-51  =  0.  15-  z2  =  24  +  4a:. 
•8.  m2  +  5  wi  -  6  =  0.  16.  7  +  2x  =  a;2. 

9.  a;2-13z  +  40  =  0.  17.  75-3x2=75«. 

10.  x2  +  6x  +  5  =  0.  18.  19  -  a  =  4 a2. 

VERBAL  PROBLEMS 

1.  If  4  is  taken  from  a  certain  number  the  result  equals  96 
divided  by  that  number.    Find  the  number. 

2.  Find  the  two  consecutive  numbers  the  sum  of  whose 
squares  equals  113. 

3.  In  physics  we  learn  that  the  distance  in  feet  which  a 
stone    thrown    downward    goes    in    a   given    time   equals    16 
multiplied  by  the  square  of  the  number  of  seconds  it  has  fallen, 
plus  the  product  of  the  velocity  with  which  it  is  thrown  and 
the  number  of  seconds  fallen ;  that  is,  s  =  vt  + 16 1*.    Suppose 
that  v  =  20  ft.  per  second  and  s  =  1800  ft.    Find  the  value 
of  t.    Try  to  state  the  meaning  of  this  problem  in  simple 
(nontechnical)  words. 

4.  How  long  will  it  take  a  baseball  to  fall  from  the  top  of 
the  Washington  Monument  (555  ft.)  if  it  starts  with  a  velocity 
of  50  ft.  per  second  ? 

HINT.  Solve  the  equation  16  t2  +  50  t  =  555. 


478  GENERAL  MATHEMATICS 

5.  How  long  will  it  take  a  body  to  fall  800  ft.  if  it  starts  at 
20  ft.  per  second  ? 

6.  How  long  will  it  take  a  bomb  to  fall  from  a  Zeppelin 
1000  ft.  high  if  it  starts  with  no  initial  velocity  ? 

7.  Two  trains  are  175  mi.  apart  on  perpendicular  roads  and 
are  approaching  a  crossing.   One  train  runs  5  mi.  an  hour  faster 
than  the  other.    At  what  rates  must  they  run  if  they  both 
reach  the  crossing  in  5  hr.  ? 

NOTE.    The  175  means  the  distance  along  the  track. 

8.  The  circumference  of  the  fore  wheel  of  a  carriage  is  less 
by  3  ft.  than  the  circumference  of  the  hind  wheel.    In  traveling 
1800  ft.  the  fore  wheel  makes  30  revolutions  more  than  the 
hind  wheel.    Find  the  circumference  of  each 

wheel. 

9.  A  window  (Fig.  312)  in  the  form  of  a 
rectangle  surmounted  by  a  semicircle  is  found 
to  admit  the  most  light  when  the  width  and 
height  are  equal.   If  the  area  of  such  a  window  p 
is  175  sq.  ft.,  what  is  its  width  ? 

10.  A  boy  has  a  piece  of  board  16  in.  square.    How  wide  a 
strip  must  he  cut  from  each  two  adjacent  sides  to  leave  a 
square  piece  whose  area  is  three  fourths  that  of  the  original 
piece  ?    In  what  form  would  you  state  your  result  to  meet  all 
practical  purposes  ? 

11.  A  lawn   is  30  ft.  by  80  ft.    Two  boys  agree  to  mow 
it.    The   first  boy  is    to    mow  one    half  of   it    by  cutting   a 
strip   of  uniform   width  around  it.    How  wide  a  strip  must 
he  cut? 

12.  A  farmer  has  a  field  of  wheat  60  rd.  wide  and  100  rd. 
long.    How  wide  a  strip  must  he  cut  around  the  field  in  order 
to  have  one  fifth  of  the  wheat  cut  ? 


QUADKATiC  EQUATIONS  479 

13.  In  a  circle  of  radius  10  in.  the  shortest  distance  from 
a  given  point  on  the  circumference  to  a  given  diameter  is  8  in. 
Find  the    segments   into    which    the 

perpendicular  from  the  point  divides 
the  diameter. 

HINT.  Study  Fig.  313  and  try  to  re- 
call the  various  theorems  we  have  proved 
involving  mean  proportional.  If  you  fail 
to  get  a  solution,  refer  to  the  mean  pro- 
portional construction  (Art.  374). 

14.  A  broker  sells  a  number  of-  rail- 
way  shares    for   $600.    A    few  days 

later,  the  price  having  risen  f  10  a  share,  lie  buys  for  the  same 
sum  three  less  shares  than  he  sold.  Find  the  number  of 
shares  transferred  on  each  day  and  the  price  paid. 

15.  A  boy  sold  a  bicycle  for  $24  and  lost  as  many  per  cent 
as  the  bicycle  had  cost  him  in  dollars.    Find  the  cost. 

16.  A  line  20  in.  long  is  divided  into  two  parts,  A  C  and  CB, 
so  that  AC  is  the  mean  proportional   between  CB  and  AB. 
Find  the  length  of  A  C. 

*538.  Maxima  and  minima  algebraically  determined.  We 
have  seen  in  Art.  531  that  the  graph  of  a  quadratic  func- 
tion may  be  used  to  determine  the  maximum  or  minimum 
values  of  the  function.  This  method  is  not  as  exact,  how- 
ever, as  the  algebraic  method  of  determining  maxima  and 
minima,  and  it  is  longer.  Take,  for  example,  the  problem 
of  Art.  531  to  find  the  maximum  value  for  20  x  —  2  a-2. 
If  we  represent  this  maximum  value  by  m,  then  we  may 
write  20  x  —  2  a-2  =  m.  If  we  divide  both  sides  of  the 
equation  by  2  and  complete  the  square  on  the  left  side 
of  the  equation, 

(x  -  5)2=  25  -|. 


480  GENEKAL  MATHEMATICS 

It  is  evident  from  this  equation  that  if  x  is  a  real 
number,  m  cannot  be  greater  than  50.  Therefore  the  maxi- 
mum value  of  m,  or  of  20  x—  2x*,  is  50. 

In  like  manner  by  letting  m  represent  the  minimum 
value  of  a  function  we  can  determine  when  in  is  a  mini- 
mum more  quickly  and  more  accurately  than  we  can  by 

the  graphic  method. 

EXERCISES 

Determine  the  maximum  or  minimum  values  of  the  follow- 
ing functions  : 

1.  3xa-4ar-l.  3.   ->x*-x-l.  5.  x*  -  6x  +  8. 

2.  2  +  2Z-2X2.         4.  1-x2.  6.  6-x-x2. 


SUMMARY 

539.  This  chapter  has  taught  the  meaning  of  the  follow- 
ing words:  parabola,  maxima  and  minima. 

540.  This  chapter  has  taught  three  methods  of  solving 
a   quadratic    equation    of    one    unknown:    the    graphical 
method,  the  factoring   method,  and    the  method  of  com- 
pleting the  square. 

541.  The  graphical  method  proved  to  be  the  most  con- 
crete. It  presented  a  clear  picture  of  the  changes  hi  the  value 
of  the  function  which  correspond  to  changes  in  the  value  of 
the  unknown.   It  also  served  as  a  sort  of  "  ready  reckoner." 

542.  The   factoring   method   was   more  expedient;  the 
results  were  obtained  by  this  method  much  more  quickly 
and  with  greater  accuracy. 

543.  The   method  of  completing  the  square  was  used 
to  solve  quadratic  equations  which  were  not  solvable  by 
the  method  of  factoring. 

544.  Both  the  graphical  and  algebraic  methods  of  de- 
termining maxima  and  minima  were  presented. 


INDEX 


Absolute  value,  156,  161 
Acute  angle,  49 
Addend,  39 
Addition,  algebraic,  162 

commutative  law,  38 

geometric,  36 

law  of,  4 

of  monomials,  160-163 

order  of  terms  in,  38 

of  polynomials,  168 

of  positive  and  negative  numbers, 

160-169 

Adjacent  angles,  60 
Ahmes,  88 
Ahmes  Papyrus,  360 
Algebra,  origin  of  word,  8 

"shorthand"  of,  13 
Alloy  problems,  338 
Altitude,  67 
Angle,  acute,  49 

bisection  of,  68 

central,  52 

complement  of,  119 

construction  of,  56,  62 

definition,  47 

degree  of,  55 

of  depression,  352 

of  elevation,  350 

initial  side  of,  48 

left  side  of,  114 

measurement  of,  54,  66 

negative,  159 

notation  for,  50 

obtuse,  49 

positive,  159 


Angle,  reflex,  49 

right,  48 

right  side  of,  114 

straight,  48,  111,  112 

supplement  of,  116,  118 

symbols  for,  48 

terminal  side  of,  48 

vertex  of,  47 
Angles,  adjacent,  60 

alternate  exterior,  127 

alternate  interior,  124 

comparison  of,  59 

complementary,  119 

corresponding,  68 

exterior,  139 

geometric  addition  of,  61 

geometric  subtraction  of,  61 

interior,  125,  130 

positive  and  negative,  159 

size  of,  48 

supplementary,  116,  118 

supplementary     adjacent,     111, 
116 

vertical,  122 
Arc,  degree  of,  65 

intercepted,  52 
Area,  measurement  of,  74 

of  a  parallelogram,  79 

practical  method  of  estimating,  74 

of  a  rectangle,  75 

of  a  rhombus,  81 

of  a  square,  78 

of  a  trapezoid,  82 

of  a  triangle,  81 

unit  of,  74 


481 


482 


(JKNKKAL   .MATIIK.MATK'S 


Areas,  calculating,  205 

proportionality  of,  341 
Arithmetic  average,  244,  ii4o,  247 
Arrangement,  167 
Ascending  powers,  167 
Axioms,  21,  22,  37 

Bar  diagram,  construction  of,  224 

interpreting,  222 
Base,  102 

Beam  problems,  336,  385 
Bearing,  of  a  line,  353 

of  a  point,  354 
Bhaskara,  108 
Binomial,  cube  of,  416 

geometric  square  of,  91 

square  of,  92 
Bisector,  construction  of,  66,  68 

perpendicular,  66 
Braces,  175 
Brackets,  175 
Briggs,  Henry,  59,  447 

Cartograms,  230 
Centigrade,  289 
Central  tendencies,  measures  of, 

244 
Characteristic,  428 

table  of,  437 
Circle,  arc  of,  52 

center  of,  52 

circumference  of,  52 

construction  of,  51 

definition  of,  51 

degree  of  arc  of,  53 

diameter  of,  52 

quadrant,  52 

radius  of,  52 

semicircle,  52 
Class  interval,  239 
Class  limits,  240 


Coefficient,  39 
Coincide,  34 

Commutative  law,  38,  85 
Compasses,  31 

measuring  segment,  32,  33 
Compensating  errors,  255 
Complement  of  an  angle,  119 
Consecutive-number  problems,  13 
Constant,  301 
Coordinates,  265,  368 
Corresponding  parts,  346 
Cosine,  357,  358,  361 
Cube,  98 

Cube   root,  of   arithmetical   num- 
bers, 420 

by  slide  rule,  456 

by  table,  397,  398 
Cumulative  errors,  255 
Curve,  normal  distribution,  257 

skewness  of,  260 

symmetry  of,  259 

Data,  214 

Decagon,  44 

Decimal  point,  logarithms,  435 

Degree,  of  angle,  53 

of  arc,  53 

of  latitude,  53 

of  longitude,  53 

of  a  number,  166 
Dependence,  300 
Dependent  variable,  300 
Depression,  angle  of,  352 
Descartes,  108 
Descending  powers,  167 
Difference,  of  monomials,  42 

of  two-line  segments,  38 
Direct  variation,  308 
Dissimilar  terms,  7 
Distance,  280 
Division,  checking  long,  209 


INDEX 


483 


Division,  definition  of,  194 
law  of  signs  in,  195 
of  monomial  by  monomial,  105 
of  negative  numbers,  194 
of  polynomial  by  monomial,  197 
of  polynomial  by  polynomial,  207 
with  slide  rule,  454 
by  zero,  211 

Drawing  to  scale,  345-355 

Elements  of  geometry,  88 
Elevation,  angle  of,  350 
Elimination,  373 

by  addition  or  subtraction,  374, 
375 

by  substitution,  377 

summary  of  methods  of,  379 
Equal  segments,  34 
Equation,  checking,  6 

definition  and  properties,  12 

laws  for  solving,  2-5,  9 

members  of,  2 

number  satisfies,  6 

quadratic,  462 

root  of,  7 

substituting  in,  6 

translation  of,  12 
Equations,  contradictory,  371 

dependent,  372 

equivalent,  372 

exponential,  443,  448 

identical,  372 

inconsistent,  372 

indeterminate,  370 

outline  of  systems,  373 

pupils'  test  of  simple,  243 

simultaneous  linear,  367,  369 

system  of,  369 

systems  containing  fractions,  381 

in  two  unknowns,  373 
Equiangular,  149 


Equilateral,  44 

Euclid,  88 

Evaluation  of  formulas,   93,  279, 

290 
Exponent,  indicates  degree,  166 

logarithm  as  an,  427 
Exponents,  102,  425 

fractional,  412 

negative,  415 

zero,  414 
Extremes,  323,  328 

Factors,  198 

common  monomial,  198 

"cut  and  try,"  199 

of  difference  of  two  squares,  202 

equal,  102 

prime,  200 

of  trinomial  square,  202 
Fahrenheit,  289 
Fallacious  proof,  211 
Formula,  definition,  78,  273 

evaluating,  93,  279,  290 

interest,  273 

motion  problem,  279 

solving,  276,  294 

summary  of  discussion,  279 

translating  into,  287 

work  problem,  283-285 
Fourth  proportional,  334 

construction  of,  334 
Frequency  table,  239 
Fulcrum,  336 
Function,  359 

defined,  299 

dependence,  300 

graph  of,  301 

graphical  solution,  304 

linear,  303 

quadratic,  463 
Functions,  trigonometric,  359 


484 


GENERAL  MATHEMATICS 


Geometric  problems,  376 
Geometry,  origin  of  word,  88 
Graph,  215     ' 

of  centigrade,  290 

of  constant-cost  formula,  261 

of  data,  214 

of  Fahrenheit,  290 

of  functions,  301 

of  linear  equations,  263-269 

of  quadratic  equations,  463-469 

of  simultaneous  linear  equations, 
369 

of  variation,  308-311 

of  x  =  10*,  442 

of  y  =  x2,  391 
Graphic  curve,  construction  of,  233 

interpreting,  231 
Graphing,  terms  used  in,  267 

Hexagon,  44 
Hipparchus,  360 
Hyperbola,  311 

Identity,  204 

Independent  variable,  300 

Index,  420 

Indirect  measurement,  345-355 

Inequality,  34 

Inertia  of  large  numbers,  255 

Intercepted  arc,  52 

Interest  formula,  273-275 

involving  amount,  277 
Interest  problem,  graphical  solu- 
tion of,  276 

solved  by  logarithms,  444 
Interpolation,  437 
Intersection,  point  of,  26 
,  Inverse  variation,  309-311 
Is8sceles  triangle,  138 

Joint  variation,  312 


Labor-saving  devices,  424 

Latitude,  63 

Law  of  the  lever,  336 

Laws,  21,  38,  85,  179,  196,  336 

Least  common  denominator,  324 

Least  common  multiple,  11 

Length,  26 

measurement  of,  27 

units  of,  28 
Lever  arm,  181 
Line,  bearing  of,  353 

of  sight,  351 
Line  segments,  26 

difference  of  two,  38 

equal  and  unequal,  34 

ratio  of,  35 

sum  of,  36 
Linear  equations,  definition,  266 

graphic  solution  of  simultaneous, 

369 

Locus,  266 
Logarithm,  of  a  power,  433 

of  a  product,  432 

of  a  quotient,  434 
Logarithms,  424,  450,  451 

applied  to  slide  rule,  449 

historical  note  on,  446 

notation  for,  427 

position  of  decimal  point,  435 
Longitude,  53 

Mannheim  slide  rule,  450,  453 
Mantissas,  428 

table  of,  430-431 
Maxima,  468-469,  480 
Mean  proportional,  329 
Means,  323,  328 
Measurement,  of  angles,  54 

of  areas,  74 

errors  in  precise,  30 

indirect,  345-355 


INDEX 


485 


Measurement,  of  line  segments,  27 

of  volumes,  99 
Median,  66,  250 

construction  of,  67 

how  to  determine,  251 
Members  of  an  equation,  2 
Metric  system,  28 

advantages  of,  29 

historical  note  on,  29 
Minima,  468-469,  480 
Minutes  of  angle,  53 
Mixture  problems,  338,  379 
Mode,  248 

advantages  of,  249 

disadvantages  of,  249,  250 
Monomial,  degree  of,  167 

factors  of,  198 
Monomials,  division  of,  195 

multiplication  of,  184 

sum  of,  177 
Motion,  circular,  284 
Motion  problems,  279,  384 

graphical  illustration  of,  283 
Multiplication,  abbreviated,  95 

algebraic,  89-90 

by  balanced  bar,  180 

commutative  law  of,  85 

law  of  signs  in,  179 

of  monomials,  184 

of  a  polynomial  by  a  monomial, 
186 

of  positive  and  negative  num- 
bers, 178-194 

several  factors,  184 

with  slide  rule,  453 

special  products,  192,  193 

of  two  binomials,  190 

of  two  polynomials,  187,  188 

by  zero,  183 

Napier,  John,  446 


Negative  number,  151 
Newton,  Isaac,  418 
Normal  distribution,  257,  268 
Notation,  for  angles,  50 

for  logarithms,  427 

for  triangles,  130 
Number,  absolute  value  of,  156 

algebraic,  151 

degree  of,  167 

literal,  6 

negative,  150-157,  178-180 

numerical  value  of,  156 

positive,  150-154 

prime,  198 
Number  scale,  170 
Number-relation  problems,  378 
Numbers,  difference  of  algebraic, 
177 

ratio  of,  35 

representation  of,  152-153 

sum  of  algebraic,  177 

Obtuse  angle,  49 
Octagon,  44 
Opposite  angles,  122 

of  parallelogram,  142 
Order  of  powers,  167 

Parabola,  467 

Parallel  lines,  construction  of,  69 

definition,  68 
Parallelepiped,  99 

volume  of  oblique,  100 

volume  of  rectangular,  99 
Parallelogram,  defined,  70 

opposite  angles,  142 
Parenthesis,  175-176 
Partial  products,  89 
Pentagon,  44 

Perfect  trinomial  square,  393 
Perigon,  48 


480 


GENERAL  MATHEMATICS 


Perimeter,  43 
Perpendicular,  65 

bisector,  construction  of,  66 
Pictograms,  214-220 
Point,  bearing  of  a,  354 

determined  by,  26 
Polygons,  classified,  44 

equilateral,  44 

similar,  318 
Polynomial,  40 

degree  of,  167 
Polynomials,  addition  of,  168 

classified,  40 

division  of,  208 

multiplication  of,  187 

subtraction  of,  173 
Positive  numbers,  151-154 
Power,  103 
Powers,  ascending,  167 

descending,  167 

table  of,  397,  398 

of  ten,  425-426 
Prime  number,  198 
Problems,  alloy,  338 

beam,  337,  385 

clock,  284 

consecutive-number.  13 

digit,  378-379 

geometric,  376 

interest,  273 

lever,  336 

mixture,  338,  379 

motion,  279,  384 

number-relation,  378 

recreation,  386-388      , 

specific-gravity,  340 

work,  285 
Product,  accuracy  of,  93 

geometric,  89 

law  of  order,  85 

monomial,  85 


Product,  partial,  89 
of  a  polynomial  and  a  monomial, 

86 

of  powers,  103 
of  two  polynomials,  187 

Proportion,  322 
beam  problems,  337,  385 
different  arrangements  of,  327 
mean  proportional,  329 
means  and  extremes  of,  323,  328 
mixture  and  alloy  problems,  338 
specific-gravity  problems,  340 
test  of  proportionality,  324,  328 

Proportional,  constructing  a  mean, 

332 

fourth,  334 
inversely,  309 

Protractor,  54 

Pyramid,  frustum  of,  98 
triangular,  98 

Pythagoras,  401 

historical  note  on,  402 
theorem  of,  397,  399,  400 

Quadrant,  52 
Quadratic  equation,  463 
Quadratic    equations,    completing 
the  square,  475-477 

factoring  method,  471-474 

graphic  method,  465 

two  solutions  of,  465 
Quadratic  function,  463 

graph  of,  463 

Quadratic  surd,  392,  406-407 
Quadratic  trinomial  square,  392 
Quadrilateral,  44 
Quotients  in  per  cents,  335 

Radical  expression,  420 
Radical  sign,  390 
large  number  under,  406 


INDEX 


487 


Radicand,  390 

Rate,  280 

Ratio,  35,  315,  345 

trigonometric,  359 
Rationalizing  denominator,  409 
Rectangle,  70 
Reflex  angle,  49 
Regiomontanus,  360 
Removal  of  parenthesis,  175-177 
Rhombus,  81 
Right  angle,  48 
Right  triangle,  135 
Right  triangles,  similar,  355 

sum  of  acute  angles  in,  137 
Root  of  an  equation,  6 
Roots,  fractional  exponents,  412 

higher,  420 

by  logarithms,  439 

table  of,  397-398 

Scale  drawings,  345-355 
Seconds,  of  an  angle,  53 
Sector,  217 
Segment,  line,  26 

unit,  27 
Semicircle,  52 

Series,  continuous  and  discrete,  237 
Signs,  151 

change  of,  171 

law  of,  179,  195 

minus,  151-152 

plus,  152 

Similar  terms,  40,  42 
Similar  triangles,  315 

construction  of,  314-316 

corresponding  sides  of,  315,  330 
Similarity,  345 

Simultaneous  equations,  367,  369 
Sine  of  an  angle,  357 
Slide  rule,  449 

decimal  point,  458 


Slide  rule,  index  of,  452 

inverted  slide,  457 

proportion  problems,  457 

raising  to  powers,  455 

runner,  454 

Solids,  geometric,  98,  99 
Solving  equations,  6 
Specific  gravity,  340 
Sphere,  98 
Square,  71 

of  the  difference,  192 

of  the  sum,  192,  193 

trinomial,  202 
Square  root,  390-397,  451 

algebraic  rule  for  finding,  396 

of  a  fraction,  408,  409 

by  graphic  method,  391 

by  logarithms,  439-440 

by  mean  proportional,  405 

memorizing,  408 

of  a  product,  406 

by  ruler  and  compass,  404 

by  slide  rule,  455 

table  of,  398 
Squared  paper,  32 
Statistical      regularity,      law     of, 

254 
Statistics,  214 

defined,  238 

historical  note  on,  270,  271 

limitations  of,  253 

use  of,  238 
Steel  tape,  349 
Straight  angle,  48 
'   Substituting,  6 
Subtraction,  algebraic,  171 

graphical,  38,  170 
Sum,    of    angles    about    a    point, 
112-113 

of  angles  of  a  triangle,  131 
Supplement,  116,  118 


488 


GENERAL  MATHEMATICS 


Surds,  addition  and  subtraction  of, 

410 

multiplication  and  division  of ,  41 1 
Surveying,  58 
Surveyor's  chain,  349 
Symmetry,  259 
Systems  of  equations,  369 

Table,  of  mantissas,  430-431 

of  roots  and  powers,  397-398 

of  trigonometric  ratios,  361 
Tangent  of  an  angle,  358 
Tape,  349 
Terms,  7 

dissimilar,  7 

order  of,  7 

similar,  7 

Test  of  proportionality,  324 
Tetrahedron,  106 
Theorem  of  Pythagoras,  397-399 
Transit,  58 
Transversal,  68 
Trapezoid,  82 
Triangle,  altitude  of,  67 

area  of,  81 

base  angles  of,  138 

base  angles  of  isosceles,  138 

defined,  43 

equilateral,  44 

exterior  angles  of,  139 

isosceles,  138 

notation  for,  130 

perimeter  of,  43 

right,  135 

sides  of,  43 

similar  right,  317 

solving,  359 

sum  of  exterior  angles,  139 

sum  of  interior  angles,  131-135 

trigonometric  area  formula,  365 


Triangle,  vertex,  47 

vertices,  43 

wooden,  136 
Triangles,  construction  of,  142-147 

similar,  314 

Trigonometric  ratios,  359 
Trigonometry,  356 
Trinomial,  40 

factoring,  199 
Trinomial  square,  202 
Turnin :  tendency,  181 

Unequal  angles,  60 

Unequal  lines,  34 

Units  of  measure,  28,  74,  99 

Value,  absolute,  156 
Variable,  300 

dependent,  300 

independent,  300 
Variation,  direct,  305 

inverse,  309 

joint,  312 

Vel  Deity  or  rate,  280 
Verbal  problem,  method  of  solving, 

16 

Vertex  of  an  angle,  47 
Vertical  angles,  122 
Vieta,  108 
Vinculum,  175 
Volume,  of  cube,  102 

measurement  of,  99 

of  parallelepiped,  100 

unit  of,  99 

Work  problems,  285,  383 

Zero,  division  by,  211 
multiplication  by,  183 


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